LIBRARY OF CONGRESS, 



3®^ 



Chap.y.i...*edpyfight No 

Shelf.„i.V_fe_2- 



•._ 



UNITED STATES OF AMERICA. 



I 



I 



ELEMENTS 



OF THE 



DIFFERENTIAL AND INTEGRAL 
CALCULUS 



Witb ^samples an& practical Bpplications 



/ 



BY 

J. W. NICHOLSON, A.M., LL.D. 

President and Professor of Mathematics 
Louisiana Slate University and Agricultural and Mechanical College 



<£ 



FER 13 J 896 . I 




NEW YORK AND NEW ORLEANS 
UNIVERSITY PUBLISHING COMPANY 

1896 



K 



^' 



Copyright, 1896, 

BY 

UNIVERSITY PUBLISHING CO. 



All Rights Reserved 



***1673 



PEEFACE. 



Ik many respects this work is quite different from any other 
on the same subject, though in preparing it there has been no 
attempt at originality beyond presenting the principles in a more 
tangible form than usual, and thus securing a better text-book 
for the ordinary student of mathematics. The aim has been 
to prepare a work for beginners, and at the same time to make 
it sufficiently comprehensive for the requirements of the usual 
undergraduate course. 

The chief distinction of the treatise is that it is based on the 
conception of Proportional Variations. This method has been 
employed as the most elementary and practical, and none the 
less rigorous or general, form of presenting the first principles of 
the subject (see the following Note). 

Differentiation and Integration are carried on together, and 
the early introduction of practical applications both of the dif- 
ferential and integral calculus, which this mode of presenting 
the subject permits, is intended to serve an important purpose in 
illustrating the utility and potentiality of the science, and arous- 
ing the interest of the student. The formulas for differentiating 
and integrating are, as a rule, expressed in terms of u and v 
instead of x, u and v being functions of x. The advantages thus 
secured are obvious. 

Among the additional features of special interest may be 
mentioned the following : (1) The treatment of dx as a variable 
independent of x (Art. 68, aud Appendix, A 3 ): (2) a rigorous 
deduction of a simple test of absolute convergency, without 
recourse to the remainder in Taylor's formula (Arts. 115 to 119) ; 
(3) an extension of the ordinary rules for finding maxima and 
minima (Arts. 140 to 143) ; (4) a chapter on Independent 
Integration (Chap. IX); (5) integration by indeterminate co- 



IV PREFACE. 

efficients (Arts. 211 to 216); (6) the introduction of turns in 
curve-tracing (Arts. 175 to 179) ; and (7) a new proof of Taylor's 
formula, which is believed to be as rigorous as, and less artificial 
than, those in general use (Appendix, A 5 ). 

In preparing the book the best available authors have been 
consulted, and many of the examples have been taken from the 
works of Todhunter, Williamson, Conrtenay, Byerly, Rice and 
Johnson, Taylor, Osborne, Loomis, and Bowser. 

I improve this opportunity to tender my thanks to Prof. 
William Hoover of the Ohio University, Prof. Alfred Hume 
of the University of Mississippi, and Prof. 0. D. Smith of the 
Polytechnic Institute of Alabama for valuable assistance in the 
reading of proofs. Their corrections and suggestions have 
relieved the treatise from various imperfections it would other- 
wise have contained. 

Further acknowledgments of indebtedness are also due to 
my colleague, Prof. C. Alphonso Smith, of the Department of 
English, who has aided me with his scholarly criticisms. 

James W. Nicholson. 
Baton Rouge, La., 1896. 



Note. — The method of Proportional Variations, which is the suggestion 
and outgrowth of work in the class-room, is believed to possess the follow- 
ing merits : 

(a) The conception is one with which the student is already familiar, 
for the principle of proportional changes is among the first that he encoun- 
ters, even in the lower mathematics. 

(b) It affords finite differentials, and, without introducing infinitesimals. 
or infinitely small quantities, or "the foreign element of time," has all the 
advantages of the differential notation. 

(c) In many cases the proportional variations (or differential) can be 
detected by inspection (see Arts. 31, 32, 35), and in all cases they may be 
deduced by the theory of limits. Hence the method has all the lucidity of 
finite differences and all the rigor of the doctrine of limits. 

(d) It is a method to which the doctrines of Infinitesimals and Kate of 
Change are easy corollaries. 

(<) In general, the form and properties of the increments of all quan- 
tities are due to proportion and acceleration, or to proportional and dispro- 
portional changes ; hence, a system of Calculus based on such changes 
adapts itself naturally to questions in Geometry, Mechanics, and Physics. 



CONTENTS. 



CHAPTER I. 

FUNDAMENTAL PRINCIPLES. 

[Pages 1 to 20.] 

Quantity. Variables and constants, Art. 1.* Dependent and indepen- 
dent variables, 2. Functions, 3. Increasing and decreasing functions, 4. 
Explicit and implicit functions, 5. Algebraic and Transcendental func- 
tions, 6. Continuous functions, 7. Notation of functions, 8. Examples. 

Increments. Increments of independent variable and function, nota. 
tion, and illustrations, 9. General formula, 10. Examples. 

Variation. Proportional variation, 11. Principles, 12 to 15. Dispro- 
portional variation, 16. Principles, 17 to 19. Composition of increments, 
illustrations, 20. 

Theory of Limits. Definition and illustrations of limits,, 21. Prin- 
ciples, 22, 23. Proof of the formula, Ay = mji -f- m<Ji? y and geometrical 
illustration, 24. 

Differentials and Accelerations. Definition and notation of differ- 
ential and acceleration, 25. Corollaries. Derivatives, 26. How the pro- 
portional variations or differential of y =/(#) may be found, 27 to 29. 

Differentials of Geometric Functions. Differentiation, 30. Plane 
areas in rectangular co-ordinates, 31. Solids of revolution, 32. Arcs of 
curves in rectangular co-ordinates, 33. Surfaces of revolution, 34. Plane 
areas in polar co-ordinates, 35. 

CHAPTER II. 

ELEMENTARY DIFFERENTIATION AND INTEGRATION. 
[Pages 21 to 42.] 

Differentiation. Rules for differentiating, 36. Differential of c or G, 
37. Differential of cv, 38. Differential of vy, 39. Differential of vyz, 40. 



* Similarly, the following numbers refer to articles and not to rages. 

V 



VI CONTENTS. 



v c — 

Differential of — , 41 ; also of — , 42. Differential of v n , 43 ; also of i/v, 44. 

y y 

Differential oi v -\- y — z, 45. Examples. 

Slope of Curves. Direction and slope of a line, 46. Direction of a 
curve at any point, 47. Slope of a curve, 48. tan 0, sin 0, cos <p, where 
<p is the angle of direction of a curve, 49. Examples. 

Integration. Definition and sign, 50. Dependent integration, 51. 
Integral of 0, 52. Integral of cdv, 53. Integral of v n dv, 54, 55. 

/ v n dv — - I v n cdv, 56. Integral of (a -\- bx n )Paf l ~ 1 dx, 57. Integral of 

die -\- dv — dz, 58. Examples. 

Problems. The problem of integration, 59. Definite value of the con- 
stant C, 60. Examples. Applications to geometry, 61. Areas of curves, 
62. Definite integrals, 63. Length of curves, 64. Areas of surfaces of 
revolution, 65. Volumes of solids of revolution, 66. 

CHAPTER III. 

SUCCESSIVE DIFFERENTIALS AND BATE OF CHANGE. 

[Pages 43 to 54.] 

Successive Differentials. Definition and Notation, 67. Why d.r may 

be treated as a constant in successive differentiation, 68. Examples. 

Liebnitz's theorem, 69. 

Rate of Change. Uniform change, 70. Variable change, 71. Ex- 
amples. Applications to geometry, 72. Given the rate of change of a curve 
to find the rates of change of its co ordinates, 73. 

Application to Mechanics. Velocity, 74. Examples. Positive and 
negative velocity, 75. Examples. Uniformly accelerated motion, 76. 
Formulas for the free fall of bodies in vacuo, 77. Problems in Mechanics. 



CHAPTER IV. 

GENERAL DIFFERENTIATION. 
[Pages 55 to 83.] 

Logarithms. Lemma, The limit of ( 1 -f- - 1 , as z approaches oo , 78. 

Differential of log a v, 79. Examples. 

Exponential Functions. The differential of a v , 80. Examples. 
Differential of y v , 81. Examples. 

Trigonometric Functions. Circular measure of angles, 82. Dif- 
ferential of sin v and cos v, 83. Sin dv = dv, cos dv = 1, 84. Differential 
of tan v, 85. d(cot v), 86. d(sec v), 87. , rf(cosec v), 88. tf(vers r). 89. 
Examples. 



CONTEXTS. VII 

Inverse Trigonometric Functions, ^(sin- 1 y), 90. ^(cos -1 y), 91. 
ditan- 1 y), 92. ^(cot- 1 #), 93. ^(sec- 1 y), 94. ^(cosec- 1 #), 95. ^(vers- 1 ?/), 
96. Examples. Differential of an arc in polar co-ordinates, 97. tan ip, 
sin ip, cos ip, where ip is the angle formed by the tangent and radius vector, 
98. 

Functions of two or more independent variables. How such 
functions may vary, 99. A partial differential, 100. Total differential, 
101, 102. A partial derivative, 103. The total derivative, 104. Examples. 
Function of Functions, 105. Examples. Successive partial differentials 

and derivatives, 106. - — — = - — — , 107. Examples. To find the suc- 
dx ay dy dx 

cessive differentials of a function of two independent variables, 108. Im- 
plicit functions, 109. Examples. Successive derivatives of an implicit 
function, 110. Examples. Change of the independent variable, 111. To 

find the successive derivatives of -^ when neither x nor y is independent, 

112. Examples. Miscellaneous Examples. 

CHAPTER V. 

SERIES, DEVELOPMENT OF FUNCTIONS, AND INDETERMINATE FORMS. 

[Pages 84 to 105.] 

Series. Definition, 113. To develop a function, 114. An absolutely 
convergent series, 115. Test of absolutely convergent series, 116. Coral- 
laries, 117 to 119. Examples. 

Development of Functions. Two formulas, 120. Taylor's formula, 
121, 122. Binomial theorem, 123. Developments of sin (y -\- x), sin x, 
cos x, cos (y -f- x), 124. Development of log (y -j- x), 125. Maclaurin's 
formula, 126. Developments of a x , e x , e, 127. Development of tan- 1 x> 
128. Examples. To find the value of 7t, 129. To compute natural loga- 
rithms, 130. To compute common logarithms, 131 . 

Indeterminate Forms. How the form — arises, 132. To evaluate 

functions of the form — , 133. Examples. Of the form — , 134. Examples. 

Of the forms X oo and oo — oo , 135. Examples. Of the forms 0°, oo °, 
and 1 , 136. Examples. Implicit functions, 137. Examples. 

CHAPTER VI. 

MAXIMA AND MINIMA. 

[Pages 106 to 121.] 

Definitions and Principles, Definitions and illustrations, 138. If 
J[a') is a max. or min. then f'(a') = or oo, 139. f(a') is neither a max. 



Vlll CONTENTS. 

noramin., if an even number of the roots off'(x) = and f(x) = go are equal 
to a', 140. f(a') is a max. or min. if an odd number of the roots of 
f[x) = and f(x) = go are equal to a', 141. Max. and min. occur alternately, 
142. 

Rules for Finding Maxima and Minima. When all the roots of 
f(x) = and go are known or can be conveniently found, 143. I. By sub- 
stituting a' — li and a' -]- Mor x, 144. II. By Taylor's formula, 145. Self- 
evident principles which serve to facilitate the solution of problems, 146. 
Examples and problems. 

Functions of two Independent Variables. Definition, 147. Con- 
ditions for maxima and minima, 148. Examples. 

CHAPTER VII. 
APPLICATIONS OF THE DIFFERENTIAL CALCULUS TO PLANE CURVES. 

[Pages 122 to 159.] 

Tangents, Normals, and Asymptotes. Equations of the tangent 
and normal, 149. Lengths of tangent, normal, subtangent, and subnormal, 
150. Examples. Lengths of tangent, normal, subtangent, and subnormal 
in polar co-ordinates, 151. Examples. Asymptote, 152. General equa- 
tion of an asymptote, 153. Relation of y to x when they are infinite, 154. 
Examples. Asymptotes determined by inspection, 155. Examples. 

Curvature. Total curvature, 156. Uniform curvature, 157. Variable 
curvature, 158. Radius of curvature, 159. Examples. Radius of curva- 
ture in polar co-ordinates, 160. Examples. 

Contact of Different Orders. Definitions, 161. When two curves 
cross or do not cross at their point of contact, 162. Examples. Osculating 
curves, 163. Osculating straight line, 164. Osculating circle, 165. 

Involutes and Evolutes. Definitions, 166. Elementary Principles, 
167. To find the equation of the evoluteof a given curve, 168. Examples. 

Envelopes. Definition, 169. The envelope is tangent to every curve 
of the series, 170. To find the equation of the envelope of a given series 
of curves, 171. Examples. 

Tracing 1 Curves. The general form of a curve, etc., 172. Direction 
of curvature, convex and concave arcs, 173. Point of inflection and prin- 
ciples, 174. Examples. 

Singular Points. Definition (etc.). a'-turns and ?/-turns, and multiple 
points, 175. To determine the positions of the singular points of curves, 
176. To determine the character of the multiple points of curves, 177. 
Examples. Tracing polar curves, 178. Examples. The character of 
multiple points often more easily determined by changing to polar co ordi- 
nates, 179. Examples. 



CONTENTS. IX 

CHAPTER VIII. 

GENERAL DEPENDENT INTEGRATION. 
[Pages 160 to 190.] 

Fundamental Formulas. Twenty-two formulas, 180. 

Reduction and Integration of Differentials. Reduction of Differ- 
entials, definition, and bow effected, 181. By constant multipliers, 182. 
Examples. 

Reduction of Differentials by Decomposition. How effected, 183. 
Elementary differentials, 184. Examples. Trigonometric differentials, 185. 
Examples. How trigonometric differentials may often be more conveniently 
integrated, 186. Rational fractions, 187. When the simple factors of the 
denominator are real and unequal, 188. Examples. Wben some of tbe 
simple factors of tbe denominator are real and equal, 189. Examples. 
Wben some of the factors of tbe denominator are imaginary and unequal, 
190. Examples. Wben some of tbe simple factors of the denominator 
are imaginary and equal, 191. Example. 

Reduction and Integration by Substitution. Irrational differen- 
tials, 192. Examples. Wben a -f- bx is tbe only part having a fractional 



exponent, 193. Examples. Wben \/a -\- bx -\- x- or \/a -\- bx — x* is tbe 
only sard involved, 194. Examples. Binomial differentials, 195. Con- 
ditions of integrability, 196. Examples. 

Integration by Parts. Fundamental formula, 197. Examples. 

Reduction Formulas. Definition, 198. Reduction formula fc* 

xP log" xdx, 199. Examples. Reduction formula for a x x n dx, 200. Examples. 

Reduction formulas for x n cosaxdx and x n sin axdx, 201. Examples. 

Reduction formula for X sin -1 xdx, 202. Examples. 

d t, fly* cl "V 

Integral of , " ,-203. Integral of -- and , 204. 

a -\-b cos x sin x cos x 

Approximate Integration. Last resort in separating a differential 

into its integrable parts, 205. Examples. Development of functions by 

exact and approximate integration, 206. Examples. 

CHAPTER IX. 

INTEGRATION CONTINUED. 

[Pages 191 to 210.] 

Independent Integration. Increments deduced from differentials, 

207. Examples. Increments as definite integrals, 208. Examples. A 

more convenient series, 209. Examples. Bernouilli's series, 210. 

Integration by Indeterminate Coefficients. Explanation and formu- 
la, 211. When h = or the integration is independent, 212. Illustra- 
tions and examples. Application to /sin" 1 x cos 71 x dx, 213. When k is not 



X CONTENTS. 

= or when the integration is partly dependent, 214. Illustrations and 
examples. Reduction formulas for binomial differentials, 215. Approx- 
imate integration and the elliptic differential, 216. 

CHAPTER X. 

INTEGRATION AS A SUMMATION OF ELEMENTS. 
[Pages 211 to 246.] 

Elements of Functions. Differentials may be as small as we please, 
217. Elements, 218. Signification of a definite integral as a sum, 219. 
Illustrative examples. When dx is not a constant, 220. Signification of a 
definite integral as the limit of a sum, 221. Illustrative example. Inte- 
gration equivalent to two distinct operations, 222. 

Application to Geometry. Length of curves, rectangular co-ordi- 
nates, 223. Examples. Polar co-ordinates, 224. Examples. To find the 
equation of a curve when its length is given, 225. Areas of curves, rect- 
angular co-ordinates, 226. Examples. Generatrix of area, 227. Polar 
co-ordinates, 228. Examples. Areas of surfaces of revolution, 229. Ex- 
amples. Volumes of solids of revolution, 230. 

Successive Integration. A double integral, 231. Definite double 
integrals, 232. A triple integral, 233. Examples. 

Areas of Surfaces. Plane surfaces, rectangular co-ordinates, 234. 
Polar co-ordinates, 235. Examples. Surfaces in general, 236. Examples. 

Volumes of Solids Determined by Triple Integration. Formula, 
237. Examples. 

Application to Mechanics. Woee, how computed, 238. Example. 

Centre of Gravity. Centre of gravity, moments, etc., 239. Centre of 
gravity of a plane area, 240. Centre of gravity of a plane curve, 241. 
Centres of gravity of solids and surfaces of revolution, 242. Examples 

APPENDIX. 

[Pages 247 to 256.] 

Differentiate functions, Ai. Another illustration of the formula 
Ay = mji -j- m-tft?, A 2 . The differential of an independent variable is, in 
general, a variable, A 3 . Another method of finding the differentials of a v 
and loga v, A 4 . A rigorous proof of Taylor's formula, A 5 . Completion of 
Maclaurin's formula. A 6 . The values of nil and m 2 in the formula 
Ay = rriih -f- ra 2 A 2 , A 7 . 



LIMITED COURSES. 

(a) The first three chapters. This course is complete as 
far as it goes, since differentiation and integration are carried on 
together. It embraces the notation, fundamental principles, and 
some of the most important applications of the Calculus. The 
student who understands elementary algebra and geometry, and 
the construction of elementary loci, should find but little dif- 
ficulty in mastering it. 

(b) The first six chapters. This adds to the former course 
the transcendental functions, development of functions, evalua- 
tion of the indeterminate forms, and maxima and minima. 

Suggestions. (1) It is recommended to omit the more 
difficult examples and problems in passing over the book the 
first time. 

(2) A 5 of the Appendix may be substituted for Arts. 116 to 
122, at the discretion of the teacher. 

xi 



DIFFERENTIAL AND INTEGEAL CALCULUS. 



CHAPTER I. 
FUNDAMENTAL PRINCIPLES. 

QUANTITY. 

1. There are two kinds of quantities employed in Calculus, 
variables and constants. 

Variables are quantities whose values are to be considered as 
changing or changeable. They are usually represented by the 
final letters of the alphabet. 

Constants are quantities whose values are not to be consid- 
ered as changing or changeable. They are usually represented 
by the first letters of the alphabet. Particular values of vari- 
ables are constants. 

2. Dependent and Independent Variables. A Depend- 
ent Variable is one that depends upon another variable for 
its value, and an Independent Variable is one that does not 
depend on another variable, but one to which any arbitrary 
value or change of value may be assigned. In the elementary 
differential and integral calculus, the independent variable is 
usually restricted to real values. 

Thus, in u = x* — 7x -f 5, v = (1 — x*f, y = log (1 + x), 
u, v and y are dependent variables, since they depend on the 

1 



2 DIFFERENTIAL AND INTEGRAL CALCULUS. 

variable x for their values; but x is au independent variable, 
since, as we may suppose, any value may be assigned to it without 
reference to any other variable. 

3. Functions. Dependent variables are usually called func- 
tions of the variables on which they depend. Hence, one vari- 
able is a function of another when the first depends upon the 
second for its value, or when the two are so related that changes 
in the value of the latter produce changes in the value of the 
former. 

Thus, the area of a varying square is a function of its side; 
the cost of cloth is a function of the quality and quantity; the 
space described by a falling body is a function of the time; every 
mathematical expression depending on x for its value, as x*, 
(3x — 7) 3 , 5x 2 — 6x -f 11, etc., is a function of x. 

4. Increasing and Decreasing Functions. An Increasing 
Function is one that increases when the variable increases, 
as (x -\- 1) 2 , 3x% log (5 -}- x); and a Decreasing Function is one 

5 



that decreases when the variable increases, as 1/10 — x 2 , -, etc. 

x 

A function of x may be increasing for certain values of x, 
and decreasing for other values. 

Thus, y = x 2 — 4x -f- 5 is a decreasing function for all values 
of x < 2, but increasing for all values of x > 2. 

5. Explicit and Implicit Functions. An Explicit Function 
is one whose value is directly expressed in terms of the variable 
and constants. 

Thus, in the equations y = (a — x) 5 , y = x* -f Zx 4- 5, y is 
an explicit function of x. 

An Implicit Function is one whose value is implied in an 
equation, but not expressed directly in terms of the variable and 
constants. 

Thus, in the equation x 2 -j- 2xy + 5y = 10, y is an implicit 
function of x, and x is an implicit function of y. By solving the 
equation for x or y, the function becomes explicit. 



FUNDAMENTAL PRINCIPLES. 3 

6. Algebraic and Transcendental Functions. One variable 
is called an Algebraic Function of another when the two 
are connected by an algebraic equation; that is, an equation 
which contains a finite number of terms involving only constant 
integral powers of the variables, or an equation which admits of 
being reduced to this form. 

Thus, in y = x 2 — 5x, or x 2 y* — xy b -f- 8xy — 5 = 0, or 
yl — Vax* + y = 7, y is an algebraic function of x, and vice 
versa. 

If two variables are connected by an equation which is not 
algebraic, each is called a Transcendental Function of the other. 

Thus, if y = sin x, y is a transcendental function of x, and 
x of y. 

The following are the elementary transcendental functions: 

A Logarithmic Function is one that involves the logarithm 
of a variable; as, log x, log {a -\- y). 

An Exponential Function is one in which the variable 
enters as an exponent; as, a x , y x . 

A Trigonometric Function is the sine, cosine, tangent, etc., 
of a variable angle ; as, sin x, cos y. 

An Inverse-Trigonometric Function is an angle whose 
sine, cosine, tangent, etc., is a variable; as, sin -1 x, cos -1 ?/, 
tan -1 t, etc., which are read, "an angle whose sine is x" "an 
angle whose cosine is y" etc, 

7. Continuous Functions. A function of a variable is con- 
tinuous between certain values of the variable (1) when it has 
a finite value for every value of the variable, and (2) when the 
changes in its value corresponding to indefinitely small changes 
in the value of the variable are themselves indefinitely small. 

Thus, in y = ax -\- b, or y — sin x, or y = e x , y is continu- 
ous for all finite real values of a:; so also in y = Va? — x 2 , but 
as a real quantity only for real values of x > — a and < a. 

x 
Again, y = is not continuous between the limits x = 1 

X rJ 

and x = 3, for when x =. 2, y = oo . 



4 DIFFERENTIAL AND INTEGRAL CALCULUS. 

8. Notation of Functions. The symbol f(x) is used to 
denote any function of x, and is read, " function of x." To de- 
note different functions of x we employ other symbols, as F(x) y 
f(x), (p(x), 0(x), etc. According to this notation, y —f(x) rep- 
resents any equation between x and y when solved for y. 

Thus, solving the equation y 2, — 2axy + foe 3 = for y, we 
obtain y — ax ± V<fx* — bx 3 , or y = f(x). 

The result of substituting any number, as m, for x in f(x) is 
denoted by f(m). 



Thus, iif(x) 


= x* - bx + 6, 




/(0) = 2 -5-0 + 6 = 6, 




/(l) = l 2 -5-l + 6 = 2, 




/(2) = 2 2 -5-2 + 6 = 0, 




/(3) = 3 2 - 5-3 + 6 = 0, 




/(4) = 4 2 -5-4 + 6 = 2, 




etc., etc. 



In f(x) if a; be increased by h the result is denoted by 

/(*+i). 

Thus, if /(») = z 2 + 5#, then 

f(x + A) = (s + /0 2 + 5(s + h) 

= x 2 + 5a; + (2a? + 5)7* + 7* 2 . 

EXAMPLES. 

1. In the function f(x) = a; 2 — 9# + 14, (I) which are the 
constants ? (2) Which is the variable ? (3) Find the values of 
/(0),/(2),/(7). (4) Which is the least: /(3),/(5) or/(6) ? 

Ans. (1) 9 and 14; (2) x; (3) 14, 0, o"; (4)/(5). 

2. Given f(x) = x 1 — ibx + 24; (1) show that /(3) ~/(7) 
= 0; (2) that /(5) </(!); (3) that /(- 1) =/(ll); (4) that 
f(x + h) = x* - lOx \- 24 + (2x - 10)7* + /*'. * 



FUNDAMENTAL PRINCIPLES. 5 

3. Reduce 2x + %y -j- 12 = to the form y = f{x). 

y =- 4(12 + 2x). 

4. Reduce x 2 + «/ a = i? 2 to the form ?/ =f(x). 

y = ± |/i2 a — ar\ 

5. Given /(a) = - 4 - fa; show that/(0) -/(3) = 2. 

6. Given f(x) = Vlmx; show that /(4m) — f{m) — 2m. 

7. It fix) = 1/100 - x\ show that/(6) =/(8) + 2. 

8. In tfy 2 -j- Vx* = a 2 b 2 , is y a function of z ? Why ? What 
function ? 

(1) It is. (2) Because any change in the value of x pro- 



duces a change in the value of y. (3) ± — Vd l — x 2 . 

ct 



INCREMENTS. 

9. If the independent variable be made to change from one 
value to another, the quantity by which it is changed is called 
its Increment; and this increment is positive or negative accord- 
ing as the variable is increasing or decreasing. 

When the independent variable receives an increment, the 
corresponding change in the value of any function of it is the 
increment of this function, and is found by subtracting the old 
from the new value of the function. Hence, if the function is 
increasing, its increment is positive; and if decreasing, its in- 
crement is negative. 

The increment of a variable is denoted by writing the letter 
A {delta) before it. Thus, Ax does not mean A times x, but 
"the increment of x," and is so read. Similarly, Au, A(x 3 ), and 
A {x 7, -f- 7x) denote the increments of u, x 3 and x 2 -f 7x. 

In this book li will generally be used instead of Ax, as it is 
more convenient ; but it should always be remembered that 
h = Ax. 

Increments and the method of finding them are illustrated 
algebraically and graphically in the solution of the following 
examples. 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Illustrations. 1. If a; is increased by h, what will be the 

increment of the function u = ex ? 

A P P The value of u (= ex) may be 

represented graphically by the 
area of the rectangle OB PA, 
whose base OB — x, and whose 
altitude OA = c. 



Aa 



X B h 

Fig. 1. 



u = cx= OBPA. 



(1) 



When x is increased by BG (= h), the new value of u will be 
c(x -f- h), or area of OOP' A. Hence, denoting the increment of 
u by Au, we have 

u+ Au=zex + ch= OCP'A (2) 

Subtracting (1) from (2), member from member, we have 

Au = ch = BCP'P (3) 

That is, the increment of ex is c times the increment of x. 

2. By how much will u = x* be increased when x is increased 
bjh? 

The value of u (= x*) may be represented / 

graphically by the area of the triangle OBP 
whose base OB = x and whose altitude BP 
= 2x. Hence 

u=x 2 = OBP. ... (1) 

When OB (= x) is increased by BO (= h) 
the new values of u, x' 2 and OBP will be, re- 
spectively, u + Au, (x -f hy and OOP' ', thus ° 
changing (1) into 

u + Au = x 2 + 2x7i + h* = OOP'. 



B C 

Fig. 2. 



(2) 



Subtracting (1) from (2), member from member, we have 
Au = 2xh + h 2 = BCP'P; 



or 



Au = 2xh + ¥ = BP xh + PDP'. 



(3) 



FUNDAMENTAL PRINCIPLES. 



(1) 




3. Find the increment of 

y — x 2 — 4:X -f- 5. 

We may regard y = x 2 — 4# -j- 5 as 
the equation of a curve APP'; then, P 
being any point of the curve, x = OB 
and y = BP. 

When OB {= x) is increased by 
BC (= 7*), the new values of y, x 2 — 4a: +5 
and BP will be, respectively, y -f- Ay, 
(x + li) 2 - 4(s + li) + 5, and CP' , thus 
changing (1) into 

y + Ay = x % + 2zA + h 2 - 4x - ±h + 5 = OP'. . (2) 

Subtracting (1) from (2), we have 

Ay = (2x - 4)A + h 2 = OP' - BP = DP'. „ . . (3) 

Cor. I. Evidently DP' will be -f- or — according as CP' is 
greater or less than BP; that is, in general, Ay will be positive 
or negative according as y is increasing or decreasing. 

10. General Formula. To find the increment of 

* = /(*) (1) 

Increasing x by h, and denoting the corresponding increment 
of u by Au, we have 

u + Au=f{x + h) (2) 

(2)-(l), Au=f(x + h)-f(x). 

EXAMPLES. 

1. Find the increment of u = 7r# 2 , or the area of a concentric 
ring whose width is h and whose inner radius is x. 

Au = 7t(2x + 7^)^. 

2. Find the increment of the cube u = a; 3 . 

Jw = 3z 2 A + 3^ 2 + 7*\ 



8 DIFFERENTIAL AND INTEGRAL CALCULUS. 

3. Find the increment of f(x) — x* — 7x -\- 9. 

f(x + h) -f{x) = (3x 2 - 7)h + ZxW + h\ 

4. Given f(x) = x 3 + 2x> '+ 9; show that 

/(a; + A) - /(a;) = (3a; 2 + 4x)h -f (3a + 2)h 2 + h\ 

5. Given f(x) = Vx; prove that 

f {x + h)-f(x)=--=t-—. 

Vx -\-h + Vx 

6. Given w = — ; prove that Au — r — — =-. 

x x —j— aj/t- 

VARIATION. 

11. Proportional Variation. One quantity is said to vary 
proportionally with, or to vary as, another when the ratio of 
the one to the other remains constant. 

The sign of variation is a. Thus, y varies as x is written 
y (xx. 

Illustrations. 1. The cost per yard of cloth remaining the 
same, the entire cost (y) varies as the quantity (x). That is, 

!f«» (1) 

2. The space (s) described by a body moving with a uniform 
velocity (v) varies as the time (t). Or 

• s oc t (2) 

3. The area (u) of a rectangle having a constant altitude 
(a) varies as the base (x). Or 

u oc x (3) 

12. Principles. I. // one quantity varies as another, one 
of them is a constant multiple of the other. 

Let y oc x, 

then — = m, a constant: 

x 

hence, y — mx. 



FUNDAMENTAL PRINCIPLES. 9 

Coe. I. The variations (1), (2) and (3), Art. 11, may be 
written, respectively, 

y = mx, where m is the price per yard of cloth ; 
s = mt, where m is the velocity of the body; 
u — mx, where m is the altitude of the rectangle. 

13. II. If one variable is equal to a constant multiple of 
another, the first varies as the second. 

Let y = mx, 

where m is a constant. Then 

y 

£- = m, or y <x x. 
x * 

14. III. If each of two quantities (u and v) varies as a, 
third (x), the first quantity (u) varies as the second (v). 

Since u oc x, u = mx; (1) 

and since v oc x, v — nx (2) 

/n . , u m m 

(1) ~ (2), - = — , or u — —v. 

v ' v ' v n n 

That is, u cc v. 

15. IV. The product of two variables does not vary pro- 
portionally with either of the variables. 

Let us suppose yx oc x; 

then yx = mx, or y — m, 

which is contrary to the hypothesis, since m is a constant. 

16. Disproportional Variation. When one quantity does 
not vary as another, the variation is said to be disproportional. 
Hence, one quantity varies disproportionally with another when 
the ratio of the one to the other does not remain constant. 

Thus, x 2 varies disproportionally with x, since x 1 -f- x (= x) 
is not a constant. 



10 DIFFERENTIAL AND INTEGRAL CALCULUS. 

17. Principles. I. The product of two variables varies dis~ 
proportionally with either of the variables (Art. 15). 

18. II. The nth power of a variable, n having any value ex- 
cept -\- 1, varies disproportionate with the variable. 

For, x n -T- x (= x n ~ l ) is a constant only when n = -f- 1. 

19. If wji vanishes loith h, mji* varies disproportionally 
with h. 

Let us suppose mji 2 oc h; 

then mji 2 = mh, or. mji = m; 

this being true for all values of h, must be true when h = 0, 
hence, since mji vanishes with h, the value of the constant m is 0. 
Therefore mji 2 varies disproportionally with h if m 2 is of such a 
character that mji vanishes with h* 

20. Composition of Increments. In general the increment 
of any function of a single variable is composed of two parts, 
one of which changes proportionally, and the other dispropor- 
tionally, with the increment of the variable. 

Thus, let y —f(x) represent any function of x, h any variable 
increment of x estimated from any particular value of x, and 
Ay(=f(x -j- h) —f(x) ) the corresponding increment of y; then 

Ay = mji -f mjf, ....... (1) 

where m 1 is constant with respect to h, and ni 2 is of such a 
character that mji vanishes with h (Art. 19). 

The proof of this property of increments will be given in 

* To say that m<Ji vanishes with h is equivalent to saying that ??i 2 does 

not involve any negative power of h, such as — , — , etc. ; for if it did, the 

value of m-Ji, when h = 0, would be finite or infinite. In general, m-Jr 
varies disproportionally with h whatever may be the character of m* , but 
the only cases we shall have occasion to consider are those in which raji 
vanishes with h. 



FUNDAMENTAL PRINCIPLES. 11 

Art. 24; we give here two examples in illustration of this im- 
portant proposition. 

1. Take the increment oiu — x 2 , which we have found to be 
Au = 2xli + h\ 

By reference to Fig. 2, it will be seen that BCDP = 2xh and 
PDP' — li\ Regarding BP (— 2x) as constant, and BO (= h) 
as variable, the first part of Au, BCDP, varies proportionally 
with li, and the second part, PDP', disproportionally with h. 

By comparison with (1), m 1 = 2x and m 2 = 1. 

2. Take the function u — x 3 — -7x -f- 9, the increment of 
which we have found to be 

Au = (3x 2 - 7)h + (3x + h)li\ 

Here, regarding x as constant, the part (3x 2 — 7)h varies as 
h (Art. 13), and the other part, (3^ + h)h 2 , changes dispropor- 
tionally with h (Art. 19). Comparing this increment with (1), 
we have 

wi i = 3x 2 — 7 and m 2 = 3^ -\- li. 

THEORY OF LIMITS. 

21. The principles of limits, in addition to other merits, 
afford an admirable method and means of finding the propor- 
tional increments of related variables. For convenience of refer- 
ence, and in order that the method of proportional changes and 
that of limits may be made to throw light upon each other, we 
give here a statement of such of the principles of limits as we 
shall have occasion to employ. 

The Limit of a variable is a constant which the variable ap- 
proaches, and from which it can be made to differ by less than 
any quantity which may be assigned, but which, on the other 
hand, it can never actually reach. 

Thus, if the number of sides of a regular polygon inscribed 
in, or circumscribed about, a circle be indefinitely increased, the 
area of the circle will be the limit of the area of either polygon, 
and the circumference will be the limit of the perimeter of 
either, 



12 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Again, lets = l-|-£ + i + i + etc. 

By increasing the number of terms of this series, the value 
of 8 approaches 2 indefinitely, but can never reach it; therefore 
2 is the limit of s. 

22. Principles. I. The difference between a variable and 
its limit is a variable whose limit is 0. 

23. II. If tivo variables are continually equal, and each 
approaches a limit, their limits are equal. 

For, let X — Y be the variables and A and B their respec- 
tive limits, and suppose X -f- x — A and Y -f- y — B, then 
x — y — A — B; but the limits of x and y are (Art. 22), 
therefore, at the limit, A — B = 0, or A = B. 

24. Proof of the Formula Ay = mji -f mji\ . (Art. 20) 
Let y —f(x) be a continuous function such that 

*y or /o* + /q-/(*o 

Ax h 

approaches a definite limit (say m,) as h approaches zero,* then 

A%i 
the value of —^- must be of the form m t -f- wi 2 7i, where w 2 /z is a 

quantity whose limit is or one which vanishes with h. 

Aii 
That is, -— = m 1 -f- mji 2 . .*. z/y = mji -f- mji ~ 

where wz, is evidently independent of h. 

Geometrical Illustration - .! Let y = f(x), as defined 
above, be the equation of a curve APm, where x = OB and 
y = BP. Let TPt be a tangent to the curve at P. When x 
is increased by BC = h, we have Ay — DP'. Draw the 
secant SPP', and we have 

^ = ^ = tanX^ (1) 

* See Appendix, Ai. f For another illustration see Appendix, A 2 . 



FUNDAMENTAL PRINCIPLES. 



13 



Now taking the limits of the two members of (1), remember- 
ing that as h approaches 0, P' , m 
approaches P, and XSP ap- 
proaches XTP as its limit, we 
have 

m = ttmXTP, 

which is a definite quantity de- 
pendent on a particular value 
{x f = OB) of x, and independent 
of h. 

Again, since m 1 = tan XTP, mji — D/?, and since 

Ay = Z>P' = mji + m 2 /* 2 , 

w 2 /^ 2 = tP', which is a quantity that vanishes with h. 




DIFFERENTIALS AND ACCELERATIONS. 

25. The Differential of a function is that part of its incre- 
ment which varies proportionally with the increment of the 
independent variable, and the Acceleration is that part which 
varies disproportionally with the increment of that variable. 

Thus, Art. 20, (1), the differential of y is mji, and the accel- 
eration of y is mji-. 

The differential of a quantity is denoted by writing the letter 
d before it. Thus, dy is not d X y, but the differential of y, 



and is so read. The differentials of functions like x c 



x* + Ix, 



and Vl 4- x\ are denoted by d(x a ), d{x 2 -f 7a?) , and d(Vl + a; 2 ). 

Similarly, the acceleration of a function will often be denoted 
by writing the letter a before it; as, ay, which is read, "the ac- 
celeration of y." 

Cor. I. The increment of a function is equal to the sum of 
its differential and its acceleration. Thus, 



dy = dy + ay. 



14 DIFFERENTIAL AND INTEGRAL CALCULUS. 

The acceleration may be positive or negative. 

Cor. II. When x and y are independent variables, Ax = dx, 
and Ay = dy. 

The independent variable may be supposed to change in any 
manner whatever; its increments are arbitrary, and are themselves 
independent variables dependent not even on the value of the 
independent variable itself, while the increments of the depend- 
ent variable depend on both the independent variable and its 
increments. This arbitrary character of the independent varia- 
ble leaves us free to make the most convenient supposition with 
reference to the manner of its variation, which is that this 
variation is uniform, or that its increments have no acceleration 
but are differentials. 

Cor. III. The differential of a function at any value is what 
its succeeding increment would be if at that value its chauge 
became proportional to that of the increment of the independent 

variable. That is, when y = f(x), (1) the limit of -j-, as h 

approaches 0, is 1, and (2) dy oc h. For example, in Fig. 4, 

DP' 

we have (1) the limit of -jrr, as h approaches 0, is 1, and (2) 

Dt oc PD. 

Since the differential of the function varies as the increment 
of the independent variable, the former will vary uniformly when 
the latter does. 

Cor. IV. The differential of a function is positive or negative 
according as the function is increasing or decreasing. 

Cor. V. In the increment Ay — mji -\- mjt~, dy = mji = 
m x dx, and ay — mji\ Hence in Fig. 4, since Ay — DP', and 
dy — Dt, we have ay = tP'. 

d u 
26. In the equation dy = m x dx, m 1 or -— is called the 

Derivative or Differential Coefficient of y with respect to x, and 
is equal to tan XTP, Fig. 4. 



FUNDAMENTAL PRINCIPLES. 15 

Aii 

27. Since the limit of -A as Ax approaches 0, = m l9 Art. 

24, and since -— ■ = m l3 we have 

limit r^n dj, _di ■ 

Jx = 0]_Jx_j dx' pd'*^-*- 

Aii 
which is read, " the limit of —^-, as Ax approaches 0, is equal 

dii 
to -j-" We are to understand by this that the ratio of the pro- 
portional variations of y and x can be found by taking the limit 
of -p. The student should note that the limits of Ay and Ax are 

not dy and dx; but the limit of -~- is equal to -—, because each 

is equal to m l3 just as ^f is equal to ^-f, because each is equal 
to f . 

28. In rinding the differential of a function it is not neces- 
sary to find the entire increment; only the part, or parts, involv- 
ing the first power of h or dx, will be sufficient, for the terms 
which involve the higher powers of h form the acceleration of 
the function. 

As an example let us find the differential of 

u = x* - bx* -f 3x (1) 

Increasing x by h, we have 

u + Au = (x + ny - 5(x -f ny + 3(« + k) 

= x* + 4:x*h -f- etc. - b(x z + ?>x 2 h + etc. + 3 (a -f A), (2) 
where the omitted terms contain higher powers of li. 

('2)-(l), tf« = (4a; 3 - 15a; 2 -f 3)dx, Ans. 



16 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



29. Cor. I. If u and v are functions of x, when x is increased 
by li the proportional increments of u and v are du and dv, and 
these are the parts of Au and Av which involve the first power 
of A. 

DIFFERENTIALS OF GEOMETRIC FUNCTIONS. 

30. Differentiation is the operation of finding the differen- 
tial of a function in terms of the differential of its variable. The 
process consists in finding the increment of the function and 
removing from it the acceleration, or in determining what the 
entire increment would be if it varied as the increment of the 
variable. 

The following important formulas are deduced at this time 
more especially for the purpose of illustrating the preceding 
principles. 



31. Differential of Plane Areas 
in Rectangular Co-ordinates. Let 
APP' be any plane curve, OB — x, 
BP = y, and area of OB PA = u; 
it is required to find the differential 
(du) of u. 

When x is increased by BC (= h), 
we have 




b c 



Fig. 5. 



Au = BCP'P = BCDP + PDP', 

which corresponds in form with Ay = mji -f- mjf, Art. 24, in 
which mji = BCDP, mjtf = PDP',. m x = BP = y, and mjf 
vanishes with h. 

Since the initial side of Au is BP (= y), the rectangle BCDP 
(= yli) is what Au would be had it varied as h; hence the area 
of the rectangle is the differential of u. 



'. du = yli or ydx. 



FUNDAMENTAL PRINCIPLES. 17 

Or thus: since the limit of BCP'P -r- BCDP, as BC ap- 
proaches 0, is 1, and since BCDP oc 2? (7, BCDP (= ?/^) is the 
differential of 0£P^4 (= %). 

JfeMotf % Ztmto. The increment BCP'P is > BCDP and 
<BCP'Q; that is, 

yJ»< Jw< (y + Ay)Ax; .'. y< ^< y + ^ 



As Ax approaches 0, the limit of Ay is 0, and that of —r- is 
equal to -=- ; hence we have 



fZ?< ,, . . du 



Hence, the required differential is equal to the value of the 
ordinate expressed in terms of x, multiplied by the differential 
of the abscissa. 

Cor. I. In Fig. 5, au, the acceleration of u, is the area of 
PDP'. 

32. Differentials of Solids of Revolution in Rectangular 
Co-ordinates. In Fig. 5, let v equal the volume of the solid 
generated by the revolution of OB PA about OX as an axis ; it 
is required to find the differential of v. 

When x is increased by h the corresponding increment of v is 
the volume of the solid generated by revolving BCP'P about 
BC as an axis; that is, 

Av = vol. generated by BCP'P 

= vol. gen'd by BCDP + vol. gen'd by PDP' 
= nifh + vol. gen'd by PDP', 



18 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Since the initial base of Av is the circle whose radius is y> 
the cylinder generated by BCDP (= nyVi) is what the incre- 
ment Av would be had it varied as h. 

dv = 7ty*h or 7ry*dx. 

Method by Limits. 

Vol. gen'd by BCDP < Av < vol. gen'd by BCP'Q, 

or ny^Ax < Av < n{y -f- Ay) 2 Ax. 

Av 
^ < ~Ax < ^ + Ay ^' 

Passing to limits, as in previous examples, we obtain 

dv = rcy^dx. 

Hence, the required differential is n times the square of the 
value of the ordinate expressed in terms of x, multiplied by the 
differential of the abscissa. 

Cor. I. In Fig. 5, av, the acceleration, is the volume gener- 
ated by PDP f . 

33. Differential of the Are of a Plane Curve in Rectangu- 
lar Co-ordinates. In Fig. 4, let s = the length of the arc AP, 

then PP f = As. Since Dt is what Ay would be had it varied 
as h, Pt is what As would be had it varied as7^; hence. Dt = dy 
and Pt — ds; and since Pt — VPD* + Dt% we have 

ds 



Vdx' + dif, or (|/l4-J^W 



Hence, the required differential is the square root of the sum 
of one and the square of the value of the derivative of y with 
respect to x, expressed in terms of x, multiplied by the differential 
ofx. 



FUNDAMENTAL PRINCIPLES. 



19 



Cor. I. The limit of the ratio of an arc of any plane curve 
to its chord is unity. 
For (Fig. 4), 

As_ 
arc PP' _ As _ Ax 

chord PP' ~ vz? + Zjf " i / 1 + (AyY 9 

the limit of which is evidently unity. 

34. Differential of Surfaces of Revolution in Rectangular 
Co-ordinates. Let s = the length of the arc AP, and 8 = the 
area of the surface generated y 
by the revolution of AP about 
OX as an axis, then AS = the 
area of the surface generated by 
the revolution of PP' {= As) 
about BC (= h). 

At P and P' draw PG and 
P'Q each equal to PP' and 
parallel to OX; the areas of the 

surfaces generated by revolving PG and P'Q about OX are 
2nyAs and 27r(y -f- Ay)As, aud evidently ^7$ lies between the 
former and the latter. That is, 













Q 

F 


yS 


P 


A 











B 

Fig. 6. 



2ityAs < AS < 2n{ij + Ay) As. 



AS 



As li approaches 0, the limit of Ay is 0, and the limit of -r- 



is equal to -=— ; hence we have 

il Sf 
2ity < -=— < 2 Try, or d# = Znyds, 



or dS = 2ny Vdx 2 + 




20 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Hence, the required differential is the product of the circum- 
ference of a circle tuhose radius is y, by the differential of the 
arc of the generating curve. 

35. Differential of Plane Areas in Polar Co-ordinates. 
Let APP' be any plane curve, and let be the pole7 OP (= r) 
P the radius vector, and . put 6 -- 
XOP and u = the area of OAP; it 
is required to find the differential 
of u. 

When 6 is increased by the 

angle POP' (= A 6), u will be 

a a increased by the area of POP' 

FlG - 7 - (= Au). 

From as a centre with the radius Oa (= 1) describe the 

arc "be (= Ad or d6), and with the radius OP (= r) describe the 

arc PD (= rdB). 

Since OP is the initial side of Au, the area of the sector POD 
is what Au would be had it varied as Ad. 

du = iOP X PD = ir'dd. 

POP' 
Or thus: since the limit of T> nn , as be approaches 0, is 1, 

and since POD oc be, POD ( = h'dd) is the differential of 
OAP (= u). 

Cor. I. The acceleration of u is the area of PDP'. 



CHAPTER II. 
ELEMENTARY DIFFERENTIATION AND INTEGRATION. 

DIFFERENTIATION. 

36. Every function may be differentiated by the principles 
heretofore established, but in practice it is better to use rules, 
which we now proceed to deduce. 

37. To differentiate a constant. Since a constant has no 
increment, its differential is 0. That is, 

dc — 0, and dC = 0. 

Hence, the differential of a constant is 0. 

38. To differentiate the product of a constant by a 
variable. 

Let u = cv, (1) 

where c is a constant and v is a function of x. 

Let h represent any variable increment of x estimated from 
any particular value of x, and du and dv the corresponding pro- 
portional increments of u and v, Art. 29; then 

u + du = c(v -\- dv) = cv -\- cdv (2) 

(2) - (1), du = cdv (3) 

.*. Rule. — Multiply the constant by the differential of the 
variable. 

3 9. To differentiate the product of two variables. 

Let u = vy, (1) 

where v and y are functions of x. 

21 



22 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Let x 9 represent any particular value of x, and u 9 , v 9 , y 9 the 
corresponding values of u, v, and y; then 

«' = ^'2/' (i) 

* Let h represent any variable increment of x estimated from 
x', and die, dv, and dy the corresponding proportional incre- 
ments of u, v, and y (Art. 29); then 

u' + du = {v 9 + dv){y' + dy) - (dv){dy) 

= Vy + y'efo 4- v'dy (2) 

The term (dv)(dy) is eliminated, or dropped from the second 
side, since it varies disproportionally with h (Art. 17) and is 
therefore a part of the acceleration of u. Subtracting (1) from 
(2) we have 

du = y'dv -f- v'dy. 

Since y' and v 9 are any corresponding values of y and v, we 
have 

du — d(vy) = vdy + ydv. 

.'. Rule. — Multiply the -first by the differential of the second, 
and the second by the differential of the first, and add the two 
products. 

40- To differentiate the product of any number of 
variables. Let u — vyz, where v, y, and z are functions of x. 

Assume w = yz, 

therefore u = viv\ 

then du = (w)dv -\- v{dw), 

also die = zdy -j- ydz; 

therefore du — (yz)dv -j- v(zdy + ydz) 

= yzdv -\- vzdy + fl^ffe. 
In a similar manner it may be shown that 

d(vtvyz) — vwydz -f- viozdy -f- vyzdw -j- wyzdv. 



ELEMENTARY DIFFERENTIATION AND INTEGRATION. 23 

.*. Eule. — Take the sum of the products obtained by multi- 
plying the differential of each by all the other variables. 

41. To differentiate a traction. 

Let u = — , where v and y are functions of x, 

y 

v 
Since u — — , we have uxi = v. 

y 

Differentiating, udy + ydu = dv. (Art. .39) 

, 7 dv dy 

TT ., dv — udy y ° 
Whence, clu = = - 

y y 

__ ydv — vdy 

~ ? • 

.". Rule. — Multiply the denominator by the differential of 
the numerator and from the product subtract the numerator 
multiplied by the differential of the denominator, and divide the 
result by the square of the denominator. 

42. Cor. I. The differential of a fraction whose numerator 
is a constant is minus the numerator into the differential of the 
denominator divided by the square of the denominator. 

For, Art. 41, dl- ) = y ° ~ CC y = — ^-, since dc = 0. 

V y y 

43. To differentiate a variable having a constant ex- 
ponent. 

Let u = v n , where v is any function of x, and n is any con- 
stant integer or fraction. 

I. When the exponent is a positive integer. 

Since v n = v . v . v . . . to n factors, 

d(v n ) = v n -\lv + v n ~ l dv ... to ^ terms (Art. 40) 

= nv n ~ l dv. 



24 DIFFERENTIAL AND INTEGRAL CALCULUS. 

II. When the exponent is a positive fraction. 

a 

Let u ~ v c , then u c = v a . 
Differentiating, cu°~ x du = av a ~ x dv. 

Substituting for m, cv c ^ = av a ~ x dv. 

a 
a- J 

Reducing, cv da = av a ~ 1 dv. 



Dividing by «v " " ' , du = jT 1 ** 



III. When the exponent is negative. 

Let u — v~ m , then u — — -. 

v m 



mv m ] 



Differentiating, Art. 42, du = ^r^ v 



Hence, for all the cases, we have this 

Rule. — Take the product of the exponent, the variable with 
its exponent diminished by 1, and the differential of the variable. 

Thus: d(x b ) = bx'dx, d(u~ 7 ) = - 7tT 8 d«, 

d(v % ) = \vHv y d(%~ 1 ) = - \z~ h dz, 



3 V7J' 



The same rule holds when the exponent n is irrational or 
imaginary. 

44. Cor. I. The differential of the square root of a variable 
is the differential of the variable divided by twice the radical. 



ELEMENTARY DIFFERENTIATION AND INTEGRATION 25 

For, d{ tfv) = d(tf) = iv^dv = —— 

2 Vv' 

45. To differentiate the algebraic sum of several 
variables. 

Let u = v -\~y — z, (1) 

where v, y, and z are functions of x. 

Let h represent any variable increment of x estimated from 
any particular value of x, and du, dv, dy, and dz the correspond- 
ing proportional increments of u, v, y, and z; then 

u + du = v + dv + y + dy — (z -f dz). . . (2) 

(2) - (1), du = dv + dy- dz. 

.-. Eule. — Take the algebraic sum of their differentials. 

EXAMPLES. 

Differentiate the following: 

1. y — x 3 + 5a; 2 - 3x + 7. 

dy = d(x 3 ) + d(bx*) - d(Sx) + d(7) (Art. 45) 

= d(x 3 ) + 5d(x*) - Sd(x) + d(l) (Art. 38) 

= 3x*dx -f lOxdx — 3dx + (Arts. 37, 43) 

= (3a; 2 + 10a; - 3)dx, Ans. 

2. y = x 2 -f 53: + 3. <fy = (2a; + 5)<fo. 

3. y = a; 4 - 4a; 3 - 3a; 2 . dy = (4a; 3 - 12a; 2 - 6x)dx. 

4. y - 3a; 5 - 7a; 2 - 9. dy = (Vox' - etc.)dx. 

5. y = x 6 + 2a; 5 - 7a; + 10. dy = {6x b + 10a; 4 + etc.)ds. 

6. y = a; 2 + a;" 2 . tfy = (2a; - 2ar 3 )<fo. 

7. y = ^" 3 + «*• dy = (- 3ar* + ^"V^ 

8. y = a; 1 - «a;- 4 + 5. dy = (fz s + 4aar 5 )da;. 

9. y " = 6a£ - 4a: 1 + 2a; 1 . dy = (21a: 1 - etc.)dx. 

10. y = 3aT* - 6x~* +5. dy = (- aT 1 -f ±x~*)dx. 

11. y = as™ - far*. tfy = (r/^rt^" 1 + bnx- n - x )dx. 

12. y = VU + 4VS. tfy = -~ + H?. 



26 DIFFERENTIAL AND INTEGRAL CALCULUS. 

13. y = fix + -. (= a;* -f a;- 1 .) o> = (4aT s - ar 8 )da;. 

X 

15. «/ 2 = 4aa\ dy — — dx. 

16. ?/ 2 + ^ 2 ^i? 2 . rfy= - ?<fc 

17. ay + &V = « 2 6 2 . dy = - h ~dx. 

Find the following: 

18. d(a?+. l)(x 2 +2x). 

(x 2 + l)rf(aj s 4- 2x) 4- (a; 2 4- 2z)<Z(z a 4- 1) (Art. 39) 
= (x 2 4- 1)(2oj 4- 2)^ 4- (a; 2 4- 2*)2a;rfaj 

= (2a; 3 4- 2a; 1 4- 2x 4- 2)dx 4- (2a; 9 4- 4o?)dz 
= (4%* 4- 6a: 2 4- 2a; 4 2)dx, Ans. 

19. d(x - l)(x* 4- x 4- 1). 3a: 2 da;. 

20. d(ax i y i ). &axly 3 dz 4- 3ax*y-dy. 

21. d(a;)(«/ 2 4- 1).- (?/ 2 4- l)Ac 4- &zy<fy. 

22. 0> 2 - l)(a: 4 4- 1). (6a; 5 - 4a; 3 4- 2x)dx. 

23. dl$x h y k ). 3x~*y h dx 4- 2x h y~ l dy. 

24. d[ar a (l 4- a" 3 )]. (- 2ar 3 - 5x «)dx. 

25. ^(14-a;)(:c + a; 2 )a;. 

26. d(l + 2x*)(l + 4x % ). 

27. a> 4- 1) (a; 3 - a; 2 4- x - 1). 4a; 3 o\r. 

28. d(x* 4- a)(3x 2 4- 6). (15a- 4 4- 36a- 2 4- 6ax)dx. 

29. o'(12a:V 4- Voabe). I8x h i/dx 4- l^ifdy. 

Differentiate the following: 

30. u = \±*> 

1 — a; 



ELEMENT AR T DIFFERENTIA TION AND INTEGRA TION 27 

du= a-*w+*)-a+*m-*) (Art41) 

(1 — x) v ' 

2dx 

, Ans. 

, ' adx 

dU = 7T~. 





-(1-aT 


31. 


a — x 


X 


32. 


l + x 

u = —-. — a . 

1 -\-x* 


33. 


X* 


11 - (i + *r 


34. 


x 3 x 1 


U - x* - 1 x-1 


35. 


x 3 
~ a 2 — x 2 ' 


36. 


2a 2 — 3 



(1 - 2a - a 2 )da 

*» = — (FT**— 

7 3 a 2 da 



du 



2xdx 



du - 8*' + 6s+12 
4a + a 2 * tf *~ (4a + a 2 )* ^ 

3 
(a + to) 3 * 

a 3 7 3a 2 4- a; 3 _ 

0+^- *• = (T+iV* 5, 



39. m = fa 2 — 3a + 5. 

d(a 2 -3a+5) (2a-3Wa /A , JJX 

dw = — . = \ - J (Art. 44) 

2Vx* - 3a + 5 2 fa 2 - 3a + 5 

40. w = VW^b. du = — I^L=. 

V 3a 2 — 5 

. . . 2 4-% 

41. u = f4a + i 7 9a 3 . dw = V fe 

2 fa 

2 -f 3a 



42. « = afl 4- a. f??t = — — dx. 

2 Vl + a 

43. w = — - dw 



Vl - a 2 (1 - x *f 

A A i/l 4" X 7 ^ 

44. ?« •= \ —± — . du = == 

1 ~ x {l-x)Vl-x> 



28 DIFFERENTIAL AND INTEGRAL CALCULUS. 

45. v = (x a -3a? a + 4a;- 5) 5 . 

du = 5 (a; 3 - 3a; 2 + 4a; - 5)W(a; 3 - 3a; 2 + 4x - 5) (Art. 43) 
= 5 (a; 3 - 3a; 2 -f 4a; - 5) 4 (3a; a - Gx -f 4)<7a;. 

46. w = (a; 3 - 7a; -f 5) 3 . dw = 3(a; 3 -7a;+5) 2 (3a; a -7)da\ 

47. u - (x 2 + 5a; - 9) 1 . du = ^(x' i -{-5x-9) i (2x-\-b)dx t 

3dx 



48. it = (2Vx+ 3)~ 3 . <fo 



Vx(2Vx + 3)*' 



, Q I x \ m 7 mx^dx 

49. w = L . dw = — . 

U-a7 (1-s) 

50. w = (« + zWa — x. du = ^- ^ ^ a? . 

2*^- a; 

51. y = - 



z(x* + ^f dx x \x 3 + lf 

ko „ _ V ( x + <0' ^ (x-2a)Vx + a 

rx-a »# (a; — a) 3 



SLOPE OF CURVES. 

46. Direction. The direction of a straight line is deter- 
mined by its angle of direction, which is the angle formed by 
the axis of a; and the line. 

The Slope of a line is the tangent of its angle of direction. 
Thus, in Fig. 4, the angle of direction of the line TP is 

XTP, and the slope of the line is tan XTP = -^ = % % 

r PD dx 

47. The direction of a curve at any point is the same as 
that of a tangent to the curve at that point. 

For, at the point P (Fig. 4), the deviation of the secant 
SPP' from the curve PP', arising from the formers cutting 
the latter, diminishes indefinitely as P' approaches P ; and 
since the tangent TPt, which touches the curve at P, is the 
limiting position of the secant, the tangent has the same direc- 
tion as the curve at the point P or (x, y). 



ELEMENT AR Y DIFFERENTIA TION AND INTEGRA TION. 29 

48. Cor. I. The slope of a curve at any point is the slope 
of its tangent at that point. Therefore the slope of a curve at 

the point (x, y) is -,-. The differential of the arc of a curve at 

any point is a straight line laid off on the tangent to the curve' 
at that point, Art. 33. 

49. The angle of direction of a curve at the point (z, y) is 
usually denoted by 0; that is, = XTP, Fig. 4. 

dy • ^ &y j ^ dx 

.'. tan = -f-, sin = -~, and cos = -=- 3 
ax ds ds 

where ds = the differential of the arc of the curve, Art. 33. 



EXAMPLES. 

1. The equation of a curve is y = x 3 — 2x; find the slope of 
the curve at the point (x,y). 

Differentiating the equation and dividing by dx 9 we have 

^ = 3x 2 - 2, Ans. 
dx 

2. In the same curve find the slope at the point where x = 1. 

dx 

3. Find the slope of the parabola y* = 9x at the point where 

x = i. ^ = 1 

dx 4 

4. In the same parabola find the point where the curve makes 
an angle of 45° with the axis of x. 

Since tan 45° = 1, we make -/■ = — = 1: hence 2y = 9, 

dx 2y * 

y- 9 

or y = U, and x = -- = — — 2£. 
v 9 4 

5. Id the circle y' 1 -f- x 1 = B 3 , find the point where the slope 
of the curve is — f. Where y — \R, and x — \R. 



30 DIFFERENTIAL AND INTEGRAL CALCULUS. 

6. In the same circle find the point where the curve is par- 
allel with the line whose equation is by + \2x = GO. 

Where x = \^R. 

7. At what angles does the parabola f = Qx cut the circle 
f + x* = 16 ? 

Find their slopes at their points of intersection; then find 
the angles between the lines having these slopes. Thus: solving 
the two equations, we find (for one of the points of intersection 
of the two curves) x = 2 and y = 2 ^3. The slope of the 

Q 

parabola at this point is — (= i ^3), and that of the circle 

is ( = — ^ 4/3). Therefore the taugent of the required 

angle is 

i^+i^ _ = i^l = 5 ^- = 8 8 8 
l - i 4/3 x i Vd 1 - i 3 

8. At what angles does the line 3y — 2x — 8 = cut the 
parabola y 2 = 8x? Tan -1 .2 and tan -1 .125. 

9. The equation of a curve is y = x z — 9x 2 -f 24x — 11. 
(1) Find the slope of the curve at the point where x = 3. (2) 
Find the values of x at the points where the slope of the curve 
is 45. (3) Find the values of x at the points where the carve is 
parallel with the axis of x. 

(1) - 3; (2) x - 7 and - 1; (3) x = 2 and \ 

10. Find the point where the curve y = x 2 — Ix -f- 3 is par- 
allel to the line y = 5x -j- 2. Where a; = 6. 

11. Find the point where the parabola if — kax is parallel 

with the circle y 2 — 2Rx — x 2 . Where x = R — 2a. 

2 2 
x y 

12. Show that the ellipse 7-5 + ^ = 1 intersects the hyper- 

lo o 

bola x 2 = y 2 -\- 5 at right angles. 

13. At what angles does the circle x 2 -f- y 2 = Sax intersect 

x 3 

the cissoid if — — — ? 

Za — x 

At the origin, 90°; at the other two points, 45°. 



ELEMENTAB T DIFFERENTIA TION AND INTEGRA TION. 31 



INTEGRATION, 

50. Integration is the inverse of differentiation, Thus, 
while differentiation is the process of deriving the differential of 
a function from the function, integration is the process of de-~ 
riving the function from its differential. The function is called 
the integral of the differential. Thus, the differential of x 3 
being 3# 2 dx, x 3 is the integral of 3x 2 dx. 

The Sign of integration is / ; thus / 3x*dx indicates the 
operation of integrating 3x 2 dx, therefore / 3x 2 dx = x 3 , which is 
read " the integral of 3x 2 dx is x 3 ." The two signs d and / annul 

each other;* / d(x 3 ) = x 3 and d I (Sx^dx) = 3x 2 dx. 

51. Dependent Integration. When the process of inte- 
grating depends on reversing the corresponding process of dif- 
ferentiating, as is usually the case, it is called dependent inte- 
gration. This process will now be employed in establishing 
rules for integrating elementary differentials. 

52. To integrate 0. 

Since dC = 0, Art. 37, we have A) — C. 

Therefore, as may be added to any differential without 
changing its value, the general form of its integral will not be 
complete without an indeterminate constant term. This con- 
stant term, as will be seen, may be eliminated, or determined 
from the data of any particular problem. 

53. To integrate the product of a differential and a 
constant. 

Since d(cv) = cdv, Art. 38, we have / cdv = c dv = cv -J- C. 
* To be exact, dfj\x)dx - f(x)dx, hwxfdfix) =f(x) + C. 



32 DIFFERENTIAL AND INTEGRAL CALCULUS. 

.'. Kule. — Integrate the differential, and multiply the result 
by the constant. 

54. To integrate v n dv, where n has any positive or nega- 
tive, integral or fractional value except — 1. 

,n+l 



I V n+1 \ 

Since d\ J = v n dv, Art. 43, we have 

/» v n+1 

J n 4-1 



.-. Kule. — Remove dv, increase the exponent by 1, and divide 
the result by the neiv exponent. 

Thus, fafdz = K + C; fy l dy = *y* + 0; 

fv-'dv = - ib~ 3 + G; fzu-tdu = 6m* + 0. 

55. Cor. I. The above rule applies also to / {v) n dv, where v 
is any function of a variable. 

Thus, fix 1 - 3a; + 5) 3 [2a; - S]dx = i(x* - Sx + 5) 4 + C. 

In this example v = x 2 — 3x -J- 5, and £?v = (2a; — 3)c?a;; that 
is, the rule applies whenever the factor without the ( ), viz., 
[2x — V[dx, is the differential of the quantity within the ( ). 

56. Cor. II. / v n do = — / (v) n cdv; introducing a constant 

thus is sometimes necessary to render the quantity without the 
( ) the differential of the one within. Thus, 

f(4x + 2a; a ) § (l + x)dx = i/* (4s + 2a; 2 ) s (4 + 4a;)c?a; 

57. Cor. III. Any differential of the form (a + bx n fx m dx, 
where n = m + 1, can be integrated by the above rule; thus, 

J [a + bx n ) p x n - l dx = Kf(a + bx n ) p nbx n - l dx 



ELEMENTARY DIFFERENTIATION AND INTEGRATION 33 

= wA a + b ^ a + ^ = =^+w + a 

Thus, y*a VZ+5*' dx = y (3 + teWxdx = -^(3 + 5z') ? + C 

fjprhy =/ (5 + 7 *TV<fc = - *(5 + w)- 3 + o. 

The rule, Art. 54, does not hold for n = — 1, for the reason 
that ^(v ) is not v~ l dv, but 0. The formula for this case will be 
derived subsequently, Art. 180, formula 2. 

58. To integrate the algebraic sum of two or more dif- 
ferentials. 

Since d(u -\- v — z) = du -+- dv — dz, (Art. 45) 

we have / (du -f- dv — dz) = I du -j- I dv — I dz 

= u -f- v — z -J- C. 
.*. Rule. — Take the algebraic sum of their integrals. 

Thus, y*(3z a - 2x + 5)dx =fdx 2 dx - j \xdx +f$dx 

— z 3 - x* + 5a: + C 

EXAMPLES. 
Integrate the following: 

1. dy = (6z 3 - 42: + 5)dx. 

y — J 6x*dx — J ixdx -f / hdx (Art. 58) 

= 6 J x\lx - 4 Jxdx -f 5 y^ (Art. 53) 

= 6-| _4.| 2 + 5-o:+(7 (Art. 54) 

= 2x 3 - 2a: 2 + 5z + C, Ans. 

2. <fy = (7x* - bx*)dx. y = x 7 -x* + C. 

3. dy — (5x 2 — Zx~ l )dx. y = f x + aT 3 -f (7. 



34 DIFFERENTIAL AND INTEGRAL CALCULUS. 

4. dy = (3x* + 2x~ 3 )dx. y = 2x i - x~* + C. 

5. dy = {x h + x l )dx. y = %x % -\- \x l + C. 

6. dy = -^—^ or x~*dx. y = f x 5 -j- £7. 

7. dy = -^ or of *fe y = — i^~ 3 + CI 



re 



8. <fy 



(rtrc 2 H = )dx. y = i^rc 8 + Vx + (7. 

\ 2 Vx* 



J \x* x 3 x*J * 3x 3 ' 

10. e?y = (1 + x'yx'dx. 

2/ =y*(l + atyiofdz = if {I + z 4 ) 3 d(l + z 4 ) (Art. 56) 

= tVU + O* + ft Ans. (Art. 55) 

11. dy = (l + x) 3 dx. y = J(l + z) 4 + (7. 

12. tfy = (1 + z) S ffo. y = |(1 + z) s + a 

13. «fy = (1 - x)~\lx. y = (1 - a)" 1 + (7. 

14. <fy = (a + a 2 )*A y = i(a + a: 2 ) 1 + C. 

15. dy = {a- x*)-*a?dx. y = - f (a - z 3 )* + C. 

16. ^ = (6 + x~*)*x~*dx. y = - J(5 + af*) 1 + (7. 

Find the following: 

17. y (z 4 - 3ar 4 )dse. \x -f a;" 3 + (7. 

18 ' / vfe = /( X ~ ^ fe ~ ^^ + a 



|/1 
19. f(x+-— ^L=)dx. 



| a + ^(l + ^) § + C. 



20. / ( - '-=■— )dx. Vax + c V» 8 + C. 

\ 2Vx J 

2x -3 > 
V2iC 3 - 6z + 5> 



21. f[- —^~ 3 W VSz 2 - 6* + 5 + G. 



ELEMENT AR T DIFFERENTIA TION AND INTEGRA TION 35 

22. yV + 3z 2 - §)\x 2 + 2x)dx. f (x s + 3:c 2 - 6) § + £ 

/* xdx 

23. / ? . 

J (1 -z 2 )*. 



25 



»/ 



6rr 2 ^ 
V^z" — 5' 






5+ (7. 




4/1-3 

£-2</ 2 




^4z 3 - 




i |/3z 2 -<6x-\- r 
x - x 3 + \x h - 


K + a 



26 . f_ £=ML_ 

J V3x* -6x + 7 

27. f{l - xydx, 

29. f( a + bx + CX \ x. 2^x(a + ibx + icx*)+C. 

v V x ' 

30. y Vz(a 2 - xydx, 2^— — + -jj- - J5 j + a 

The following differentials may be reduced to the form of 
(a -f lx n Y% n -Hx, and then integrated by Art. 57. 

r x~*dx__ = /» ar 3 ^ V^T^ 3 

^ |V + ic a ^ j/a^ar 8 + 1 " a 2 z + 

Multiply the binomial under the radical sign by x~ 2 , and the 
numerator of the fraction by x~\ 

00 I" Vx T ^~a 1 dx (^ - tff , n 

32 ' J ^r 3«V "^ °* 

QQ P V2ax - x 2 dx (2ax-x 2 ) § 

34. /_^_ 3 . *=+a 

t/ (a 2 -hz 2 )* a*Va* + x* 

35. /* ^ ^ 1 (7 

' ^ (2az - z 2 ) 1 « Vaaa: - tf 3 



36 DIFFERENTIAL AND INTEGRAL CALCULUS. 



PROBLEMS. 

59. The Problem of Integration is the inverse of that of 
differentiation; if the latter is " given a curve to find its slope," 
the former is "given the slope of a curve to find the curve"; 
if the latter is " given a function to find its rate of change," 
the former is " given the rate of change of a function to find 
the function." Since the general problem of differentiation is 
"given a function and its variable to find their proportional 
changes," that of integration is "given the proportional changes 
of a function and its variable to find the function." 

60. Definite Value of the Constant C. To complete each 
integral as determined by the preceding rules, we have added 
a constant quantity C. While the value of this constant is un- 
known, it is said to be indefinite ; but it becomes definite when 
its value is assigned or determined by the conditions of the 
problem under consideration. The signification of C and the 
manner of determining its definite value are illustrated in the 
following examples. 

EXAMPLES. 

1. Required the equation of the curve whose slope at the 
the point (z, y) is 4:e 3 — 2x + 3. 

By Art. 48, the slope of a curve at (x, y) is -—. 

or dy = (4z 3 — 2x + $)dx. 

Integrating, y = x* — x* -f- 3x + 0, 

which is the indefinite integral or equation required. 

To determine the value of C we may make x = 0, which 
gives y — C, where y indicates what y becomes when x = 0, 
and is therefore the distance from the origin to where the curve 
cuts the axis of y. 



ELEMENTARY DIFFERENTIA TION AND INTEGRATION. 37 

Therefore the equation may be written 

y = x* — x 1 + Zx + # n , 

which becomes definite when y is known. 

When we know any corresponding values of an integral and 
its variable, C can b? determined. Thus, if it is given that the^ 
last curve passes through the point x' = 2, y f == 10, then when 
x — 2, y — 10, and we have 

10 = 2 4 - 2 2 + 3 2 + C, or C = - 8. 

2. What is the equation of a curve which passes through the 
origin, and whose slope at the point (x, y) is (a -f- x)* ? 

civ 
Here ~= (a -\- a;) 2 , or dfy = (a -j- jc) 9 fl?#. 

Integrating, y = J(a -f #) 3 -f (7. 

Since the curve passes through the origin, y — 0; hence, 
making x = 0, we have 

and the required equation is 

v = H a + x Y - i« 3 . 

3. Required the equation of a curve which passes through 
the point {%' = 3, y' = 11), and whose slope at (x, y) is 3,c 2 — 
10a; + 1. y - x* - 5x' + x + 26. 

4. The differential of an integral is (3 -f- x^xdx, and the 
integral is when x — 0; required the value of C. 

C=- 1/3. 

5. The differential of a function is (1 -f \xfdx, and the 
function is when x — 0; find the function. 

¥ \(i + I*) 1 - ¥ v 

6. A function and its variable vanish simultaneously, and 
their proportional changes are as x^a* -f x/)^ to 1; find the 
function. |(« 2 _|_ x *f _ |^ 

Take rfa; for the increment of x, then x^a* + z 3 )~^£ will be 
the proportional increment of the function. 



38 DIFFERENTIAL AND INTEGRAL CALCULUS. 

7. Find the area of a plane curve whose differential is 
V±ax dx, and whose value = when x = 0. \x V±ax. 

8. AYhen x is increased by dx the proportional increment of 
the arc of a certain curve is (9 -f- x*)*x*dx\ find the length of 
the arc, supposing it and x to start from the same point. 

|(9 + x'f - 4± 

9. The area of a surface of revolution is increased propor- 
tionally by 7r(l -f- x — x*)~^(l — 2x)dx when x is increased by 
dx; find the area of the surface, if it = when x = 0. 

[(1 + a; - a^)* - l]2ar. 

10. The differential of the volume of a solid of revolution is 
7r(2Rx — x*)dx; find the volume supposing it and x to vanish 
simultaneously. 7t[Rx 2 — £#']. 

61. Applications to Geometry. The last four problems 
indicate the possibility of finding the length and area'of plane 
curves and the area of surfaces and volume of solids of revolu- 
tion when their differentials are known. Now Arts. 31, 32, 33, 
34 enable us to find the differentials when the equation of the 
curve is given. 

62. Areas of Curves. 

11. Find the area of OBPA, the equation of the curve APE 



■N 


i 


3 


3 




*/ 


being y — x* — Sx -f- 15, 
where x = OB and y — 
BP. 

By Art. 31, the differ- 
ential of OBPA (= u) is 


a b i 


Fig. 8. 






x ydx, aud in the piesent ex- 




F 


ample y — x* — 8a; + 15. 


H( 


mc 


e, 


die = 


: ydx 


= (*■ 


- 8x + 15)dx; 



.-. u = t f(x" - 8a; + 15)dx = ix s - 4a; 2 + 15a; + C. 
Evidently the area u = when x = 0, hence C = 0. 



ELEMENT AB Y DIFFERENTIA TION AND INTEGRA TION 39 

63. Definite Integrals. If, in any indefinite integral, two 
different values of the variable be substituted, and the one result 
subtracted from the other, C will be eliminated, and the result 
is called a definite integral. 

Thus, Fig. 8, if Oa — 1 and Ob = 2, to obtain the area of the 

section abed we substitute 2 for x in — — 4^ 2 -j- 15# ~\- C and 

o 

get 16f -f- C (= area of ObcA), and then substitute 1 for x and 
obtain 11-J -\- C (— area of Gad A)-, subtract the latter from the 
former and we have 5^ (= area of abed). 

The following is the notation by which these operations are 
indicated : 



16}-lli = 5i. 



/V ~ 8a?+15)<fe= T| 3 — 4f 2 + \6x + C 

In general, / is the symbol of a definite integral, and indi- 
cates (1) that the differential following it is to be integrated; 
(2) that b and a are to be substituted successively for the variable 
in the indefinite integral; and (3) that the second of these re- 
sults is to be subtracted from the first. The operation is called 
integrating between the limits a and b. 

12. In the last example find the area of EFG. 

Since the roots of x 2 — 8x -j- 15 = are 3 and 5, OF = 3 
and OG = 5. Hence we have 

= - U = area of EFG. 



£ydx = [ 



\ - ±X 3 + lDX + C 



The result is negative because the area lies below the axis of x. 

13. In the same example find the area of OEA. 

f\lu = 18. 

t/O 

14. Find the general area of the curve y = Sx* — 2x -f 3 
estimated from the origin. £ ydx = ^_ % , + ^ 

15. Find the areas of the positive and negative surfaces 
enclosed by the curve y = x* — x and the axis of x. 

£ydx = \; £ydx=-i. 




40 DIFFERENTIAL AND INTEGRAL CALCULUS. 

16. Find the entire area of y* = (1 + x^x* between the 
origin and the point whose abscissa is x. |(1 -f- # 8 )* — t- 

17. Find the positive area of the parabola y* = ±ax. \xy. 

18. Find the area of y 2 — x\a 2 — x*) between the limits 

64. Lengths of Curves. 

19. Find the length of the arc OP, 
I the equation of the curve OP being 

y* = ax 3 . 

By Art. 33, the differential of OP 

(= s) is f 1 + ggJW To apply this 

x 

to the present curve, we differentiate 

y 1 = ax z , and obtain 2ydy = 3ax*dx. 
dy _ Sax 2 (dy\ 2 _ 9a V _ 9ax 

Hence > dx~^f [dxj ~~^j r ~ir t 

= |/l + i^Jdx = l/l + 9 -^dx = tift+toZdx. 

20. The differential of the equation of a certain curve is 
dy = Vx* — 1 dx; find the length of the curve, beginning at the 
origin. 

Here | = <&=!; • •• (g)'+l = .«; . = ^ 

21. Find the length of the arc of a curve whose equation is 
y = §(# — 1) § , measured from the point where x = 1. fa; 2 — f. 

22. Find the length of a curve the differential of whose 
equation is dy = \ f x* -f 2xdx, beginning at the origin. 

fcP + x. 

x z 1 

23. Find the length of the curve y = — -4- ■-, between the 

limits (1), x = 2, # = 3; (2), x — a, x — h. 

m i- (») h *~ a * i h ~ a 



r/s 



ELEMENTARY DIFFERENTIATION AND INTEGRATION. 41 

24. Find the general length of the curve y = f 1 — ^W%, 
estimated from the origin, / x \ ,- 



65. Areas of Surfaces of Revolution. 

25. AP is the arc of the circle 2/ 2 = i2 2 —£ 2 ; A 
find the area of the zone generated by re- 
volving AP about the axis OX. 



By Art. 34, dS = %ny\ 1 + l-p) dx; i 



V: 



(dy 



this example, since y* = B? 
Hence, 



dy 



' dx 




dS 



= 27ty]/ 



1 + — a dx = 2nRdx. 

y 



■/. 



.-. S = 2?rE / dx = ZttRx. 

26. In ex. 25 find the area of the zone generated by the 
arc PP' ', supposing OB = a and OE — I. \gf — 27iR(h — a\ 

' a 

27. In ex. 25, find the surface of the generated sphere. 

[S] + _l= 4R*\ 

28. Find the surface of the cone which is generated by 
revolving the line y = mx about the axis OX. 

[Sf= ?ryVx 2 + y\ 

29. Find the surface of the paraboloid, the generating 
curve being the parabola y* = 4ax. 

[Sf= frVa [(« + *)• -(«)»]. 

30. Find the surface generated by the revolution of the 
curve y = ax 3 about the x axis. x |~(1 + 9a V)' — 11 



42 DIFFERENTIA^ AND INTEGRAL CALCULUS. 

66. Volumes of Solids of Revolution. 

31. Find the volume generated by revolving OBPA about 

the axis OX, the equation of 
the curve AP being if = 'Sz 2 
- 36z + 105. 

By Art. 32, civ = ny*dx\ in 
this example y 2 =3x 2 — 362+ 105, 




hence 



Fig. 11. dv = 7t(3x 2 — 362 -f 10b) dx. 

,-.v =7i f\sx 2 - 36z -j- im)dx = tt(x* - 18x 2 -f 105.?;). 

* 

32. In the same example find the volume of the solid 
generated by the revolution (1) of OEA; (2) of EGF. 

(1) [ v f= 200tt; (2) [v] = - 4tt. 

33. Find the volume of the cone which is generated by 

revolving about x the triangle whose base is x and whose altitude 

is y (= mx). r x . „ 2X 7 ttw/V Jx\ 

J v ' v = 7t J (mx)dx = — - — = ny 2 - J. 

34. Find the volume of the solid which is generated by 
revolving y = x* — 4 about OX. 

v = 7tf(x* - 8a; 2 + 16)^ = 7z[\x" - \x* + 16a?] -j- C. 

35. In ex. 34, find the volume generated by the negative 
area of the given curve. r -i+ 2 __ 34 _ 2 n 

3G. Find the volume of the spherical' segments generated 
(1) by OBPA and (2) by BEP'P, Fig. 10; also find (3) the 
volume of the sphere (see Ex. 26). 

(1) [v]=7c(R*a-ia>); (2) |>f= 7t[R\b - a) - W~a*)]; 




+R 



(3) [v]_ b = %nR\ 

37. Find the volume of a prolate spheroid, the generatrix 
being the ellipse a 2 y % = a 2 6 2 — £V. M + " = knob 2 



CHAPTER III. 

SUCCESSIVE DIFFERENTIALS AND RATE OF 
CHANGE. 

SUCCESSIVE DIFFERENTIALS. 

67. The differential of any function, as u = f(x), denoted by 
du, is the first differential. The differential of the first differ- 
ential, viz., d(du), denoted by d 2 u, is the second differential. 
The differential of the second differential, viz., d(d 2 tt), denoted 
by d 3 ti, is the third differ ent i al ; and in general the differential 
of d n ~ l u, denoted by d n u, is the nth differential. 

du, d 2 ti, d 3 u, etc., are the Successive Differentials of it. 

68. If x is the independent variable, its differential (dx) is 
altogether independent of #,— the increment given x at any in- 
stant being entirely independent of the value which x may have 
at that instant: It at once follows, therefore, that the differential 
of dx with respect to x, like the differential of any other variable 
which is independent of x, is 0. Hence d(dx) = d' 2 x = 0. 

For example, take the function u — x 3 ; then 

(1) du = 3x^X1 

(2) d\i = d^x^dx + Zx'didx) = 6xdx'; 

(3) d 3 u = d\§x)dx* + 6xd(dx") = 6dx 3 ; 

(4) d*u = d(6)dx 3 + 6d{dx 3 ) = 0. 

Therefore, in finding the successive differentials of a func- 
tion, we treat the differential of the independent variable as a 
constant. See Appendix, A 3 . 

The student should observe the difference between expres- 
sions like d 2 y, dy*, and d(y' i ); d 2 y is the second differential of y, 
dy' 2 is the square of the differential of y, and d(y*) is the differ- 
ential of y 2 , or 2ydy, 

43 



44 DIFFERENTIAL AND INTEGRAL CALCULUS, 

If u = f(x), the successive derivatives or differential coef- 
jicients ox w with respect to sc are -7-, -r-^, y-j, etc. 

The successive derivatives of /(#) are also denoted by f'(x), 

f"(x),f"(x),...r(x). 

Therefore, when u = f(x), we have 

£=/», £=/», ••• S -/-(*)• 

EXAMPLES. 

1. Find the successive differentials ofy = x\ 

dy = 4x*dx. 

d*y = d(±x*dx) = 4tdxd(x % ) = \2x*dx*. 
d z y = d{12x i dx i ) = 12da?d(x*) ^ 2ixdx\ 
d'y — d(24xdx*) = 24dx 3 d(x) — 24dx\ 
d b y = d(2±dx A ) = 0. 

2. Find the successive derivatives of x % — 4x 2 -j- 3x — 5. 
Let /"(#) = a; 3 — 4a; 2 -f 3a; — 5 ; 

then /'(a?) = j (x* - 4a; a + 3x - 5) = 3a; a - 8a; + 3; 
f'(x) = ~ (3a; 2 - 8x + 3) = 6x - 8; 

/"(*)=J;(<to-8) = 6; 

/ y (*) = 0. 

3. Find the successive differentials of u = 2a; 3 — 7a;\ 

d^ = (6a; 2 — 14x)dx; d*u — (12a; — 14)tfa; 2 ; d 3 w = 12dx*. 

4. w = x* — 3a; 3 + 5a;. 

5. ^ = ar 3 . 

^ = — $%-*dx; cVu = 12ar 5 da; a ; d 3 w = — 60arW; etc. 

6. u = x~ 2 — x~ x -\- 5. 

du == — (2ar 3 — a; -2 )^; r? a w = (6ar 4 — 2x~ 3 )dx; etc. 

7. u = (s + 1)*. 

du = |(a; +!)-%#; dV. = - f(a + l)~fe 2 ; etc. 

8 - If A*) = ^ show that/'"(s) - - ^JL^, 



SUCCESSIVE DIFFERENTIALS AND BATE OF CHANGE. 45 

9. If /(*) = g-L- 4 -, show that/*(l) = ^ . 

10. If f(x) = 1=£ show that /"(I) = \. 

11. If fix) = ~^ 9 show that/"'(0) = - |. 

69. Leibnitz's Theorem. To find the successive differentials 
of the product of two variables. 

Let u and v be functions of x\ then 

t?(?^) = udv -f vdu (1) 

Differentiating (1), regarding u, v, du, dv, as functions of x, 
we have 

d 2 (uv) = ud*v + dudv -f- dvdu -\-vd 2 u t 

or d' 2 (nv) = «*Pv + %dudv + v^*w (2) 

Differentiating (2), we get 

tf 8 (ttv) = wrf't; + 3dud 2 v -f- Zd\idv + v*f w, 

from which we see that the law of the coefficients is evidently 
the same as in the binomial formula. 

RATE OF CHANGE. 

70. Uniform Change. When a variable changes uniformly 
it experiences equal changes in equal intervals of time, whatever 
the magnitude of these intervals. 

Thus, we may suppose the line ah (= x) to be generated by 
a point moving over each of the equal distances ab, be, cd, etc., 



abed k 

Fig. 12. 

in a second of time. If ab = be = cd . . . = h in., the rate of 
change of x is h in. per second. Denoting the number of seconds 

* |4, called factorial 4, means 1x2x3x4. In general, 
\n - 1 X 2 X 3 ... X n. 



46 DIFFERENTIAL AND INTEGRAL CALCULUS. 

by t, we have x = lit. Differentiating this equation, regarding 
h as constant, we have 

dx 
dx — lidt, and — — li. 
at 

Therefore, I. In uniform change the increment of the vari- 
able varies as the increment of the time. 

dx 
II. The derivative -jj- expresses the rate of change of x per 

second, when t represents seconds. 

71. Variable Change. When a variable does not change 
uniformly, its rate of change is ever changing, and is measured 
at any instant by what the change or increment would be during 
the next unit of time if it (the rate of change) were to remain 
unchanged during that time. 

Thus, when a train leaves a station it usually goes, for some 
distance, faster and faster; that is, its rate of change or velocity 
is constantly accelerated. If at any instant during this time we 
say " It is going at the rate of 20 miles per hour," we mean that 
it would travel that distance the next hour if its present rate 
should remain unchanged, or, as in uniform change, if the in- 
crement of the distance should vary as the increment of the 
time, that is, uniformly. 

The same is true of any other variable; that is, u being a 
function of time (t), the rate of change of u is measured by 
what Au would be in the next interval of time if it varied as 
At or dt; but du is what Au would be if it so varied, Art. 25, 
hence the differential of u measures its rate of change. 

Cor. I. Since du oc dt, du = mdt, 

du 

where m f = -j J is the rate of change of u per second, and, be- 
ing a function of t, is in general variable. See Appendix, A a . 



SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE, 47 

EXAMPLES. 

1. Required the rate of change of the area of a square. 
Let x = the side and u the area; then 

u = x 2 , and du — 2xdx. .'. -— = 2x-^-. 

dt dt 

That is, the rate of change of the area \-jr) is equal to the 

fdx\ 
rate of change of the side f^r ) multiplied by twice the side (2a;). 

We may omit dt and regard du and dx as the measures of 
the rates of change of u and x. Thus, du = 2xdx signifies that 
the area is increasing in square inches 2x times as fast as the 
side is increasing in linear inches. 

2. In the function u = x 2 — 4a; -j- 5, (1) at what rate is u 
increasing when x is 5 in. and increasing at the rate of 3 in. per 
second ? (2) At what rate is x increasing when it is 10 in. and 
u is increasing at the rate of 40 in. per second ? (3) What is the 
value of x at the point where u is increasing 10 times as fast 
as x ? 

Differentiating the given function, we have 

du = (2x - 4)dx. ........ (1) 

In this equation we have three quantities, viz. : du, the rate 
of increase of u; dx, the rate of increase of x; and x; hence, 
when either two of these are given the third may be found. 

(1) x = 5 and dx = 3; substituting in (1) and we have 

du = (10 - 4)3 = 18, Ans. 

(2) du = 40, x = 10; substituting in (1) and we have 

40 = (20 — 4)dx, whence dx = 2|, Ans. 

(3) du = lOdx; substituting in (1), we have 

10dx= (2x — ±)dx, or 10 = 2x — 4, whence x = 7, Ans. 

3. If x increases uniformly at the rate of 5 inches per 
second, at what rate is u = x 3 — 4# increasing when x = 10 
inches ? 1480 in. per second. 



48 DIFFERENTIAL AND INTEGRAL CALCULUS. 

4. In the function y = 2x* -\- 6, what is the value of x at 
the point where y increases 24 times as fast as x ? x = ± 2. 

5. If the side of a square increases uniformly at the rate of 3 
inches per second, what is the length of the side at the time the 
area is increasing at the rate of 20 sq. inches per second ? 

Let x = the side, u = the area; then u = x*. 

6. In the last example, supposing the area to increase uni- 
formly at the rate of 10 sq. inches per second, at what rate will 
the side be increasing when the area is 22 sq. inches ? 

5 

Take x = Vu> — 7= in. per second. 

1/22 

7. A circular plate of metal expands by heat so that its dia- 
meter increases uniformly at the rate of 2 inches per second; 
at what rate is the surface increasing when the diameter is 5 
inches ? 

Let x = the diameter, u = the area; then u = -x*. 

bn sq. in. per second. 

8. In the last problem, if the surface increases uniformly at 
the rate of 50 sq. inches per second, at what rate will the diam- 
eter be increasing when it becomes 5 inches ? 

20 . 

— m. per second. 
7t r 

9. The volume of a spherical soap-bubble increases how 
many times as fast as the radius ? When its radius is 4 in., and 
increasing at the rate of i in. per second, how fast is the volume 
increasing ? 

Let x = the radius, u — volume; then u = ^7tx 3 . 

(1) 4:7tx 2 times as fast. (2) 32^r cu. in. per second. 

10. A ladder 50 ft. long is leaning against a perpendicular 
wall, the foot of the ladder being on a horizontal plane x ft. 
from the base of the wall. Suppose the foot of the ladder to be 
pulled away from the wall at the rate of 3 ft. per minute. 

(1) How fast is the top of the ladder descending when x = 14 ft.? 

(2) How fast is it descending when x = 30 ft ? (3) What is the 
value of x when the top of the ladder is descending at the rate 



SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. 49 

of 4 ft. per minute ? (4) And what at the time the bottom and 
top of the ladder are moving at the same rate ? 

Let y — the distance from the base of the wall to the top of 
the ladder; then y = V2500 — x>- 

(1) J ft. per minute. (2) 2% ft. per minute. (3) x = 40 ft. 
(4) x = 25 V% ft. 

11. What is the value of x at the point where x 3 — 5z 2 -j- 17:t 
and x 3 — 3.r change at the same rate ? x —2. 

12. Find the values of x at the points where the rate of 
change of x* — 12^ 2 -f- 45.T — 13 is zero. x = 3, x = 5. 

13. In a parabola whose equation is y 1 = 12a', if x increases 
uniformly at the rate of 2 in. per second, at what rate is y in- 
creasing when x = 3 inches ? 2 in. per second. 

14. In the same parabola, at what point do y and x vary at 
the same rate ? When y = 6. 

15. In the ellipse whose equation is 44f# 2 -{- 25# 2 = 1111|, 
at what point of the curve does y decrease at the same rate that 
x increases ? When y = 3 and x — 5^. 

16. Find the points where the rate of change of the ordinate 
y = x* — Gar + ox -j- 5 is equal to the rate of change of the slope 
of the curve. Where x = 1 and x = 5. 

17. Two straight roads intersect at right angles; a bicyclist 
travelling the one at the rate of 10 miles per hour passes the 
intersection 2^- hours in advance of another travelling the other 
road at the rate of 8 miles per hour. At what rate were they 
separating (1) at the end of H hours after the first man passed 
the intersection? (2) At the end of 2-J- hours? (3) Eeqnired 
the distance (y) between them when it is not changing. 

(i) % = 5tV; (2) % = 10; (?) y = W fS mi. 

72. Applications to Geometry. The rates of change of 
the areas and lengths of carves and of the areas and volumes of 
surfaces and solids of revolution are given by Arts. 31, 32, 33, 
34. The applications are made as in Arts. 62, 64, 65, 6Q. 



50 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Rate of Change of Curves- Formula, ds = y 1 -f \-f-\ dx. 

18. Find the rate of change of the arc of the parabola 
y* = lax. ds — y 1 -j — dx. 

19. In the previous example, if a = 9, and x increases at the 
rate of 12 inches per second, at what rate will the arc be increas- 
ing when x = 16 inches ? ds = 15 in. per second. 

20. Show that the rate of change of the arc of the circle 

I I , DS '■• 7 R(lX 

x^ + if— IF is ds = - 

21. In a circle whose radius is 20 in., the abscissa changes 
at the rate of n in. per sec; at what rate is the arc increasing 
when x = 12 in. ? \n in. per second. 

73. Given the rate of change of the arc of a curve (ds), 
to find the rates of change of its co-ordinates x and y. 

Take the parabola y 2 = lax, then ydy = 2adx; between this 
equation and ds' 2 = dx* -\- dy 1 eliminate (1) dx and (2) dy, and 
we have 

dy = — — - ds and dx — y ds. 

Vla^-t-tf r a-x-x 

22. In the parabola y' ==■ Ix, if the arc increases uniformly 
at the rate of 5 inches per second, at what rates are y and x in- 
creasing when x — 9 inches ? 

i |/10 and | ^10 inches per second. 

23. If the arc of the circle x 1 -\- y 1 = 100 increases at the 
rate of 5 inches per second, at what rates are y and x changing 
when x = G inches ? — 3 and -f- 4 in. per second. 

APPLICATION TO MECHANICS. 

74. Velocity is the rate of change of the distance described 
by a moving body. Hence, if s = the distance, v == the velocity 
and t — the time, we have 

I. v = -rrl .'. s — I vdt. and t — I — . 

dt «/ J v 



SUCCESSIVE DIFFERENTIALS AND BATE OF CHANGE. 51 

Again, denoting the rate of change of the velocity by a', we 
have 



II. a' = ^r 



dv d 2 



dt dt 



■i > 



,-. v = I a'dt, and t — I — . 



EXAMPLES. 



1. If 6* = 3f, what is the velocity and its rate of change ? 

ds 
Since s — 3f, —- — Qt = v, the velocity; 

dv 
and since v = 6t, —rr = 6 = #', the rate of change of v. 

Clo 

Thus, if the unit of s is one foot, and the unit of t one 

ds dv 

second, v = — [= Qt] ft. per second, and a f = -jy [=6] ft. per 

clt civ 

second. 

2. A body passes over a distance of ct* in t seconds; find v 
and a', (1) in general, and (2) at the end of 9 seconds. 

(1) — *— and C —\ (2) — and - — . 

2 Vt 4 Vf W 6 108 

3. A body after moving t seconds has a velocity of 3f -f- 2^ 
ft. per second; find its distance from the point of starting. 

s =J*vdt = ?+ t\ 

4. The velocity of a body after moving t seconds is 5f ft. 
per second; (1) how far will it be from the point of starting in 3 
seconds ? (2) In what time will it pass over a distance of 360 
feet? (1) 45 ft.; (2) 6 sec. 

5. A body moves from A, and in t seconds its velocity is 14/^ ft. 
per second; (1) how far is the body from A ? (2) In how many 
seconds will the body have gone 847 feet ? 

(1) If ft.; (2) 11 seconds. 

75. The velocity is positive or negative according as s is in- 
creasing or decreasing, and a' sustains the same relation to v; 



52 DIFFERENTIAL AND INTEGRAL CALCULUS. 

therefore, if s increases as the moving' body goes forward and 
decreases as it goes backivard, the body is moving forward or 
backward according as v isiwsitive or negative. 

6. A train left a station and in * hours was at a distance of 
\tf — 4* 3 4- 16^ 2 miles from the starting-point; required the 
velocity and its rate of change, also when the train was moving 
backward, when the velocity or rate per hour was decreasing, 
and the entire distance travelled in 10 hours. 

s = it* — 4* 3 -j- 16* 2 — the distance from station. 

j = f - 12* 2 + 32* = v = the velocity. 

dv 

— = 3*' - 24* + 32 = a' = the rate of change of v. 

Civ 

The roots of * 3 — 12*' + 32* = are 0, 4 and 8; therefore v 
is negative, and the train was moving backward from the 4th 
to the 8th hour. 

Again, the roots of 3f - 24* + 32 = are 1.7 - and 6.3 -f ; 
hence a' is negative, and therefore v was decreasing, from the 
1.7th to the 6.3th hour. 

The roots of It* - 4f + 16* 2 = are 0, 0, 8 and 8; that is, 
s = when * = 8; hence the train was at the starting-point at 
the end of 8 hours, having gone backward as far as it had forward. 

Since the train moved forward the first four hours, then 
backward the next four hours, and then forward, the entire dis- 
tance passed over in 10 hours was 



M - M + M = 64 + 64 + 100 = 228 (miles). 



7. A train left a station and in * hours was moving at the 
rate of t 3 — 21f -j- 80* miles per hour; required (1) the distance 
from the starting-point; (2) when the train was moving back- 
ward; (3) when its rate per hour was decreasing; (4) when the 
train repassed the station; and (5) how far it had travelled 
when it passed the starting-point the last time. 



SUCCESSIVE DIFFERENTIALS A ND RATE OF CHANGE. 53 



(1) s =fvdt =yV - 21*' + 80t)dt = it* - 7f -f 



±0t* 



(2) From the 5th to the 16th hour; (3) from the 2.27th to 
the 11.72th hour; (4) in 8 and 20 hours; 

5 16 20 

(5) M - \s] + [si = 4658^ miles. 

5 16 

8. A traveller left a point A at 12 m., and in t hours after 
his rate per hour was 5 — t miles; (1) how far forward did he 
go? (2) At what times was he 8 miles from A ? (3) What were 
his rates per hour when at a distance of 10J miles from A ? 

(1) 12| miles; (2) 2 p.m. and 8 p.m.; (3) +2 mi. and - 2 mi. 
per hour. 

76. Uniformly Accelerated Motion is that in which the 

rate of change of the velocity («') is constant. That is, v changes 

Av dv 
uniformly or v oc t; hence, Art. 70, — - = -— = a' = the rate of 

change of the velocity. 

Formulas. v =Ja'dt = a't + C = a't + v , . . (1) 

and s =J vdt =J (a't + v )dt=ia't* +v t + s , . (2) 

in which v and s represent the initial velocity and distance; 
that is, the values of v and s when t = 0. 

If v — and s = when t = 0, then (1) and (2) become 

/2s 

— and v — V2a's. . (3) 

77. The increment of v, or acceleration, produced by gravity 
is about 32.17 ft. per second, and is usually represented by g. 
Hence, by substituting g for a' in (3), we obtain the four for- 
mulas for the free fall of bodies in vacuo near the earth's sur- 
face. When the bodies are not in vacuo the formulas generally 
are slightly inaccurate, on account of the resistance of the at- 
mosphere. 



54 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



PROBLEMS. 
1. A rifle-ball is projected from in the direction of Fwith 
a velocity of b ft. per second; required its path, knowing that 
its velocity in t seconds along the action-line of 
gravity (OX) will be gt ft. per second. 

Let OX and OF be the axis of x and y, 

respectively; then -~ = b, and — = gt; 




dt 



x, 



y — M, . . . (1) and x = \gt\ . . (2) 
Eliminating t between (1) and (2), we have 

Fig. 13. J g 

that is, the path of the ball is an arc of a parabola. 

2. A body starts from 0, and in t seconds its velocity in the 
direction of OX is 2abt, and in the direction of OY is a?f — b*; 
find its velocity along its path Onm, the 
distances in the direction of each axis and 
along the line of its path, and the equation 
of its path, the axes being rectangular. 

Let v X9 v y and v s represent respectively 
the velocities in the directions of the axes 
x and y and the path s. Then 




>. x = J2abtdt = abP; . 
\ y =yV^ - ^) (U = 4* 

s=f{aH 2 +byit = lan % -\-bn. . . . . 



Vt\ 



aH* 



(1) 
(2) 
b\ 

(3) 

Now to find the path of the body, we eliminate t between (1) 
and (2) and obtain 



CHAPTER IV. 

GENERAL DIFFERENTIATION. 

LOGARITHMS. 



78. Lemma. The limit of \l-\ — ) ,as z approaches infinity, 

1_ . 1 . 1 



is the sum of the infinite series 2+-jx- + -r^+T7 + ^ c 



Assuming the binomial theorem for positive integral values 
of z, we have 

-+'+W-!-)+i( i -J)( i -:-)+"- 



which evidently approaches 2 + -— + — - -f etc., as z approaches 

If. I- 
infinity. The sum of this series is represented by e. It is the 
base of the natural system of logarithms, which is equal to 
2.718281, approximately. 



mit (i+iy=, 

- CO \ % ) 



limit 
z — 



79. To differentiate the logarithm of a variable. 

Let u = log a v, where v is a function of x. 

55 



56 DIFFERENTIAL AND INTEGRAL CALCULUS. 

When x is increased by h we have 

An — log a (v + Av) — log a v 

(v + ^ V 
= W — J 



Au 

~Av 



i 

= v M 1 + tJ 



Passing to the limit, remembering that as h approaches 0, 
— approaches 0, and —r- approaches go , we have, as in the pre- 
ceding lemma, 

du 1 , ' .dv * 

^7 = v lo Sa e> °r du = (log a e)~-. 

Rule. — Divide the differential of the variable by the variable 
itself, and multiply the quotient by the constant log a e. 

The factor log a e, usually represented by m, is called the 

modulus of the system whose base is a. When a — e the mod- 

dv 
ulus is unity and we have du = — , simply. Herein lies the 

advantage of the natural system of logarithms, whose base is e, 
in all discussions of a theoretical nature. Hereafter when the 
base of a given logarithm is not indicated, it is to be understood 
that the base is e. 

EXAMPLES. 

1. Differentiate u = log(x* — 2x + 5). 

_ d(x* - 2x + 5) __ (3 % a - 2)dx 

x 3 - 2x + 5 ~ x 3 - 2x + 5* 

* For another method, see Appendix, A\ , Cor. III. 



GENERAL DIFFERENTIATION. 57 



VI — xl 1 — x 



2dx 1 —x 2dx 

X 



(1 -xf 1+x \~x % 

3x 2 dx 



3. u = log t^ 3 - a\ du =. _ 



4. m = log a (5z 2 — x 3 )\ du — 

fta v ' bx — x 



5. w = log (x + ^1 + a 2 ). ^ = 

6. w = log [(& — x) \/a + »]• ^ = 



_ 4m (10 — 3x)dx 
dx 



(a + 3#)^# 



7. y = \og*x. dy — 3log 2 #— . 



8. y = log 4 (log x). dy 



x 
4 log 3 (logx)dx 



x log X 
9. y = log (log a:). rf y ^ =__-.. 

dx 



10. 2/ == log (x-\-a-\- i/2ax -\- x*). dy 



\ f 2ax -j~ # a 
EXPONENTIAL FUNCTIONS. 



80. To differentiate u = r? r . 
Passing to logarithms, we have 

log u — v log a. 

Differentiating, — = dv log #. 

Multiplying by u, du = u log a dv, 

Substituting a v for u, du = a v \oga dv* 



* For another method, see Appendix, At. 



58 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Hence, the differential of an exponential function with a 
constant base is equal to the function itself into the natural 
logarithm of the base into the differential of the exponent. 

Cor. I. d(e v ) = e v dv, since log e = 1. 

EXAMPLES. 
1. Differentiate y = m VI +^\ 

- dy = m ^log (m)d( V l+^) = ™ V ^ }?*&)*& 

p \Og Xrfrr 

2.y = e l °e*. dy = - — . 

x 

3. y — e x log x . dy — e x log *(log x + l)dx. 

e x 7 e x x dx 

4- y = — • ^ = (jq^-r 

5. y — e x (l - x'). dy = e x (l — Zx* — x')dx. 



l+x 


e x (l - x 3 ). 


lOSf . 


6 1 + 6 X 


e* — e~ x 



, dx 

6. y = log — — -. <fy 



1 + 



,x 



81. To differentiate u = y v . 
Passing to logarithms, we have 

log u — v log y. 
Differentiating, — = dv log y -f v~. 

Multiplying by u, du — u log ydv -f- wy— . 

Substituting ?/ for u, du = «/ v log ydv -f vy v ~ l dy. 

Hence, the differential of an exponential function with a 
variable base is the sum of the results' obtained by first differen- 
tiating as though the base were constant, and then as though 
the exponent were constant. 



GENERAL DIFFERENTIATION 59 

The method of differentiating a function by first passing to 
logarithms, as in the two preceding demonstrations, is called 
logarithmic differentiation. It may be used to great advantage 
in differentiating many exponential functions and those involv- 
ing products and quotients. 

EXAMPLES. 

1. Differentiate y = (x* + l)^ 1 . 

dy = (x* + 1)* +1 log (x* + l)d{x + 1) + (x + l)(tf + l)*d{i* + 1) 
= (x 2 + l)*[(ar + 1) log (s» + 1) + 2(z 2 + ar)]fe 

2. y = ar*. ofy = ar^log a; -f l)t7a;. 

x/— 1 , ^'z (1 — log x)dx 
o. y — Vx or a;*. dy = * -j- - . 

e* — 1 . 2e* 7 

y e x + 1 (^ + I) 2 

5. y = e**. d?/ — e^l + log x)x x dx. 

Make m = x x , differentiate, and replace u and du by their 
values. 

6. y = ar**. tfy = a** Q + log a; + log 2 &W<fo. 

TRIGONOMETRIC FUNCTIONS. 

82. Circular Measure of Angles. If v = the length of 

the circular arc BP, and r — the length of 

the radius OB in terms of the same unit, the 

v 
ratio - is the circular measure of the angle 
r 

BOP. When r — 1, the measure of the an- 
gle is simply v, which is the length of the 
arc. This method of measuring an angle is 
called the circular or analytic system, as distinguished from the 
degree or gradual method. 




60 



DIFFERENTIAL AND INTEGRAL CALCULUS. 




83. To differentiate sin v and cos v. 

With the radius OB (= 1) describe the circle whose centre 
is 0, and let the angle BOP or its measuring arc BP be any 
value of v, which is a function of x, then 
PE = sin v and OE = cos v. 

Let us suppose BP or v to receive an in- 
crement. Take PT, a part of the tangent 
line at P, for the differential (dv) of the arc 
BP (Art. 48), then the proportional incre- 
ments of sin v and cos v will be GT and — GP, 
respectively. Therefore GT — d(sin v), and 
- GP = d(cos v). 
The angle GTP = BOP = v; hence in the triangle GTP 
we have 

(1) GT = cos # Jv, .*. e?(sin v) = cos v dv. 

(2) 6rP = sin v dv, .*. d(cos v) = — sin v <:Zv. 

Hence, the differential of the sine of an angle is the cosine of 
the angle into the differential of the angle. 

The differential of the cosine of an angle is minus the sine 
of the angle into the differential oj the angle. 

84. Ooe. I. 2^= sin (v + dv) and OF ^ cos (v -f dv), 
which approach, respectively, PI 7 and OB as v diminishes, and 
when » = 0we have sin dv = dv and cos dv = 1. That is, the 
sine of the differential of an arc is the differential of the arc 
itself, and the cosine of the differential of an arc is 1. 

85. The differential of the tangent of an angle is equal to 
the square of the secant of the angle into the differential of the 
angle. 

sin v 



For, 



tan v 



d tan v = 



cos v 

cos vd sin v 



sin vd cos v 



cos 2 v 

(cos 2 v -f- sin 2 v)dv „ 7 

^ — ' — = sec v dv. 

cos 2 v 



(Art. 41) 



GENERAL DIFFERENTIATION. 61 

86. TJie differential of the cotangent of an angle is equal 
to minus the square of the cosecant of the angle into the differ- 
ential of the angle. 

i 

For, cot v 



tan v 



d tan v ~ sec vcl v 

d cot v — — t — -„ — = r — „— = — cosec v dv. 

tan v tan v 

87. Tlie differential of the secant of an angle is equal to the 
secant of the angle into the tangent of the angle into the differ- 
ential of the angle. 

For, sec v 



cos v 



, d cos v sin v dv 
a sec v = — = „ — = sec v tan v dv. 

cos 2 V COS 2 V 

88. The differential of the cosecant of an angle is equal to 
minus the cosecant of the angle into the cotangent of the 
into the differential of the angle. 

For, cosec v = 



sin v 



, d siTiv cos v dv 

.'. a cosec v = --= — - = r-= — = — cosec v cot v dv. 

sin v sin v 

89. d vers v = d(l — cos v) = sin v dv. 

EXAMPLES. 

Differentiate the following: 

1. y = sin (x* — x). 

dy = cos (x* — x)d(x s — x) = cos (of — x)(2x — 1 )dx. 

2. y — tan 4 (x 3 ). 

dy = ± tan 3 (V)tf(tan x s ) = 4 tan 3 (z 3 ) sec a {x 5 )d(x 3 ) 

= 12 tan 3 (x 3 ) sec 2 (^ 3 ).t 2 ^. 



62 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



3. y — cos (ax). 

4. y = cos 3 x. 

5. y = sin a; cos x. 

6. y = tan 2 5a\ 

7. y = sec 3 a\ 

8. y — sin a; tan a;. 

9. y — (x cot a) 2 . 

10. «/ = a* sin x + cos re. 

11. y — x — sin a; cos x. 

12. ?/ = tan x — x. 

13. ?/ = sin (cos a). 

14. y = sin a; — J sin 3 a;, 

15. y = J cos 3 a — cos a\ 

16. y 



tZ?y = — a sin (ax)dx. 

dy = — 3 cos 2 a- sin a; ^a. 

tfy = (cos 2 x — sin 2 a;)^a;. 

dy — 10 tan (5a:) sec 2 (5a) dx. 

dy — 3 sec 3 x tan a; dx. 

dy = (sin a; + tan a sec a;)da;. 

dy = 2x cot a(cot x — x cosec 2 a)$r. 

dy = x cos a? ^/a;. 

dfy ±= 2 sin 2 x dx. 

dy = tan 2 x dx. 

dy = — sin x cos (cos x)dx. 

dy = cos 8 a; dx. 

dy = sin 3 a; dx. 



^ tan 3 a;— tan ar-j-a:. dy = tan 4 a; tfa. 

„ sin x 

x 



-\- log x cos a; 



f/a. 
dx. 



i. y — ar"~. t/?/ = x 

18. # = (sin a;) 3 '. dy = (sin a;)*[log sin x + a cot a 
Prove the following by differentiating both members (see 

Art. 50) : 

19. / cot xdx — log (sin x) -f- C. 

20. / — tan x dx — log (cos x) -f C. 

21. / sec a cosec x dx — log (tan x) -J- C'. 

22. I — sec x cosec x dx — log (cot a;) + (7. 

23. / tan a: rfa; = log (sec x) + (7. 

24. J — cot x dx = log (cosec a-) -f £7. 

/sin xdx , , x , /7 

j- ^ „„„ „. = log (vers x) + C. 



25. 

26. 

27. 



cos a- 



/ sin x cos 3 a efa = \ sin 2 a; — i sin 4 x -\- C. 
I sin 8 a: cos 3 x dx = £ sin 4 x — J sin fi a: + C 



GENERAL DIFFERENTIATION 63 

/'sin 3 x , , , ^ 

28. / — o—dx = sec a; + cos a; 4- (?. 

«/ COS £ ' 

Prove the following: 

29. If /(a) = Or - 6z + 12)e x ,f'"(x) = x*e x . 

30. If f(x) = e-^cos »,/ iv (a) = — 4e~ x cos x. 

31. If /(z) = tanz,/'"(z) = 6 sec 4 x - 4sec 2 z. 
*« T » 7 cos ic cos 3 # ^f?/ . 3 

33. If /(a) = z 3 log a:,/ iv (l) = 6. 

34. If /(«) = log sin x,f'"(^} = 4. 

35. If /(a) = sin x, /'(0) = 1; /'"(0) = - 1. 

36. If/(*) = log (1 + x),f"(0) = - 1; r(0) = (4. 

37. If f{x) = a x ,f n (0) = log" a. 

38. If/(o;) = e*log^/ iv (c) = 



i , 4 6 , 8 6 

log C + 5 + s j 

& c c c c 



INVERSE TRIGONOMETRIC FUNCTIONS. 
90. To differentiate sin -1 y. 
Let v = sin -1 y, then y = sin v. 



^ = cos v dv = Vl — y 2 dv. 

dv = , or ^(sin -1 y) = 



Vl - y 2 Vl-y* 

91 . To differentiate cos" 1 y. 
Let v = cos -1 y, then y = cos v. 



dy = — sin v c?i> = — Vl — ^ a rfv. 
tfa; = -^ , or f?(cos -1 v) = —'!— =, 

Vi - if Vi - y 

which always has the sign of — sin v. 

* To avoid the double sign ±, we shall suppose < v < — ; for any 
other quadrant the sign will be that of cos v. 



64 DIFFERENTIAL AND INTEGRAL CALCULUS. 

92. To differentiate tan -1 y. 
Let v = tan -1 y, then y = tan v. 

^ = sec 2 v dv — (1 -j- ?/ 2 )^. 

dv = ^y, or d(tnn- 1 y)= ^ , - 
In a similar manner we find 



93. ^(cofc- 1 : ?y) = 

94. ^(sec 1 y) = 

95. rf(cosec _1 2/) = 
95. ^(vers* 1 ?/) = 



dy 



y Vtf - 1 
dy 



V2y - y 9 

EXAMPLES. 
Differentiate the following : 

\. y — sin -1 Vx. 



dx 



d Vx 2 Vx dx 

dy~ 



Vl~(VxY Vl-x 2Vx-a? 
,., = tan-.(^). 

J \ - x \ 2dx 

d _ Vl + J & + '*)" <** 



"*" Vi-hJ 



+av (1 + z) 2 

3. ?/ = sec -1 wz. «?/ = — 

x yn*a? — 

. /CX -j tlJu 

4. it — ver -1 — — . dy — ■ 

J 9 * V$x-a? 



Vl - X* 

dx 



GENERAL DIFFERENTIATION. Q5 

5. y = sin -1 (3a; — 4# s ). dy = 

6. y = sin -1 (2a; — 1). dy = 

yx — x 2 ' 

¥. y = sin -1 (sin x). dy = dx. 

8. y = sin -1 (Vsin a;). *??/ = |(Vl -f- cosec a;)^a;. 

n , _j 2a; , 2<fc 

9. y = tan x - 2 . rfy 



1 - x* 1 + a; 3 



lO.^tan^fA^) *=*fa 



11. y = (a; + 1) tan -1 Vx — \/x. dy = tan -1 t/jc<fc. 

13. y = i log — | + i tan- 1 x. dy = 1 _^ 4 . 

Prove the following by differentiating both sides: 

13. A-t^-i = 1 tan- 1 - +6 

14. / = -sec x — \- C. 

J x Vx 2 - a 2 a a 



. f dX = s jn- 1 -+g r . 
J Va 2 -x* a 

/dx , x , 

— — = vers -1 — \- 
V2ax -x 2 a 



17. J V¥~=tfdx = x J^p^ + a l S i n -i * + a 

18. /" Jl^ ^-^+jg. 4/2^^^- + fevers- 1 - + (7. 

19. / — = m sin M ! — 4- C. 

<-> |/ c « _ ^ _ 2^ _ fl v \ -o J 

20. r Wfl gg _ logte+6+Vc u +(^+aa;)'+a 

21. / a — —T5-T — j = m tan M + C. 

^ cfx 4- 2rtto + ft 2 -j- c 2 \ c I 



66 DIFFERENTIAL AND INTEGRAL CALCULUS. 

r m{ad-bc)dx _ J ax'+b \ 
' J (ax + b){cx + tf) ~~ g \cx -\-d)^ 

The last four equations may be conveniently employed in 
integrating a certain important class of differentials, of which 
the following are illustrations: 

23. Eequired the integral of 



x* - 3x - 28 



Here x* — 3x — 28 = (x — 7)(x -\- 4); hence we may inte- 
grate by formula 22, thus : Make ax -{-b = x— 7, ex -f d = x -{- 4, 
and m(ad — be) = 5; we then have a = 1, b = — 1, c = 1, d = & 
and m = T 5 T ; hence, substituting in 22, we have 

/hdx 5 , (x — 7\ , ~ 

a --to-28 = n log U+lj + a 

Zdx 
24. Eequired the integral of 



4z 2 + 3a? + 1 

Integrating by Ex. 21, we have a 2 = 4, 2ab = 3, b* + £ a = 

/> 

and m#c = 3 ; whence a = 2, J = f, c = i V'7 and wi = — — , 

V7 

f Zdx 6 f -i f 2x + t \ I •* 

•V4x 2 + 3a; + l~W IT^"/" 1 " 

25. Find the integral of 



Vl6 - 12a? - 4a; 2 



Integrating by Ex. 19, we have a? — 4, 2ab = 12, c 5 — 5 2 = 16 
and ww = 5 ; hence a = 2, & = 3, c = 5, wi = |. 

/» 5^ 5 . . /2a? + 3 \ , ' ~ 

.\ / — r: — ■ = - Sill ^ + C. 

J Vl6 - 12a? - 4a? 2 2 V o / 



/• dx 2 , _ x 2a? + 1 

C dx 1 .8!^ , - 



GENERAL DIFFERENTIATION. 



67 



J 2x l - 4a; - 7 

■■/■ 



V2 . 2a; 



2 - 3 V2 



2 + 3 Vi 



+ G 



30 



32. f 



dx 
dx 



. , 3 + 2x „ 

sin" 1 ,_ + C. 

4/13 



Vm 4- ?&r + rx* 

— log (z Vr + 



rfz 



Vm + nx — rx* 
dx' 



Vr 



n 

2 Vr 

sin' 



Vm + nx + rx'J 



■■>- 



- 



\rx 



\/±mr + w 



+ c. 



m + nx + r# 3 ' 



V4wir — w a 



tan" 



/ 2rx 



\V\.mr — n 2 



0. 



97. To find the differential of an arc in polar co-ordinates. 

Let AP (= s) be the arc of a curve, the pole, OP {— r) 
the radius vector, and PT a tangent 
to the curve at P. 

Let 6 = XOP and ip = OPT. 
Increase by POP' (=40), then 
arc PP' = Js and OP' = r 4- Jr. 
Draw PD perpendicular to OP', 
then PD = r sin J0 and PP' = 
r 4- 4r — r cos z/#. 

The chord PP' = VPD 2 + DP", 




chord PP' _ a/(PD\\ ( dp '\\ 

Aa f \ AH I 1 Ad I 



Ad ~ f \Ad J ' \ Ad I 

Passing to the limit, remembering that as A 6 approaches 0, 

,, .. .. . chord PP' ,&mA0 .. DoN . A . 

the limits of — - — p^- and — ^— (Art. 33), also of cos Ad, is 
arc Jr ± 

each unity, we have 

ds =. 



J0 



r 2 + 



rfr 
tf0 



1 



^0. 



68 DIFFERENTIAL AND INTEGRAL CALCULUS. 

98. Cok. I. As P' approaches P, the angle OP'P approaches 
the angle OPT (=*/>); therefore 

A , limit \~ PD~] rdd , rdd , dr 



FUNCTIONS OF TWO OR MORE INDEPENDENT VARIABLES. 

^°. A function of two variables, as u = x 2 y -f- y % and 
u = sin (x + ?/)> is represented by f(x, y); and similarly f(x,y, z) 
represents a function of the three variables x, y and z. 

Since x and y are independent of each other, the function 
u =f(x, y) may change in three ways. Thus, let u = xy = area 

of <9£ZM, where 0£ = x and 0^4 = 
?/. (1) x may change and y not, 
which would give du = BCDP = 
ydx; (2) y may change and x not, 
which would give du = APFG = 
acty; (3) both # and ?/ may change, 



d# 



dec 



FlG ig which would give (understanding by 

du the portion of the increment 

which is of the first degree in dx and dy) du — ydx -\- xdy. 

Hence du may have three different values, and it is desirable to 

employ a notation by which they may be represented. 

100. A Partial Differential of a function of two or more 
variables is the differential obtained on the hypothesis that only 
one of its variables changes, as ydx and xdy in the previous 

example, and these are denoted respectively by y dx and -j—dy, 

or d x u and d y u. 

101. Total Differential. In a function of two or more in- 
dependent variables, if each variable receives an increment, that 
portion of the corresponding increment of the function which is 
of the first degree with respect to the increments of the variables 
is the total differential of the function. 



GENERAL DIFFERENTIATION. 69 

102. Prop. The total differential of a function of two or 

more independent variables is the sum of its partial differentials. 

Let u represent any function of x and y. 

When x becomes x -\- h, the part of the corresponding incie- 

dn 
ment of u which involves the first power of h or dx is j—dx. 

ax 

Hence, omitting the terms involving the higher powers of dx, 

Art. 29, the new value of u is 

, du .. 

u -\- -r-dx. 

dx 

In this new value of u when y is increased by h (= dy) the 

du 
parts of the corresponding increments of u and —dx which in- 

ax 

volve only the first power of dy are -j—dy and -z-[ — dx\dy; 

hence, omitting the terms involving the higher powers of dy, 
the second new value of u is 

, du -. , du , , d (du 7 \ ., 

u+ dx dx+ a y d y + Ty\ix dx ) (1 y> 

which result is the same as if x and y had changed simulta- 
neously, for the result of increasing x by dx and y by dy is evi- 
dently the same whether the changes be made separately or 
simultaneously. Hence, since tHe last term involves the product 
of dx and dy, we have 

7 du 7 . du , 

du= ix dx+ ^ d y- 

In a similar manner it may be shown that the theorem is 
true of functions having any number of variables. 

Cor. I. The total differential of a function is the sum of 
those parts of its increment which vary as the increments of the 
variables, respectively. 

Cok. II. The theorem is also true of functions whose variables 
are not independent of each other. 



70 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Cor. III. The total differential of a function of two or more 
variables, as x, y and z, is the sum of the differentials obtained 
by first differentiating as though x only were variable, and then 
as though y only were variable, and then as though z only were 
variable. 

103. A Partial Derivative of a function of two or more 
variables is the ratio of the partial differential of the function to 
the differential of the variable supposed to change. 

104. The Total Derivative of a function of two or more 
variables, only one of which is independent, is the ratio of the 
total differential of the function to the differential of the inde- 
pendent variable. 

To prevent confusion w r e shall distinguish the total differen- 
tial or derivative of two or more variables by enclosing it in 
brackets. 

Thus, if u =f(x, y), [du] = C ^dx + ^-dy, 

du du 

where j—dx and -r-dy are the partial differentials with respect 

(XX (XiJ 

to x and y, respectively. 

EXAMPLES. 
Find the total differential of — 

1. u . = z 2 — Zxy -f 2y*. 

j—dx — (2x — 3y)dx; -r-dy = — (3x — iy)dy. 
.-. [du] = {2x - 3y)dx - (3z - ±y)dy. 

2. * = *+*. [du] = 2 _M^lJ^i. 

x-y (x — y) 

. ,x r 7 .. ydx — xdy 

3. u — sin" 1 -. [du] = J1 — / -. 

V y S/if - x 2 

y xdy — ydx 



GENERAL DIFFERENTIATION. 



71 



5. u — log y x . 

6. u = y Binx . 



[du] = —dy-\- log ydx. 
[du] = y sin x log y cos x dx -f- 



sma; 



f 



105. Function of Functions. If u — F(y) and y — f{x), u 

is indirectly a function of x through y. In such cases the value 

du 
of — may be obtained by finding the value of u in terms of x, 

and differentiating the result; but it is often more easily found 
by the formula 



du _ du dy 

dx ~ dy dx 



(i) 



That is, the derivative of u with respect to x is equal to the 
derivative of u with respect to y multiplied by the derivative of 
y with respect to x. 

Thus, if u = tan -1 y, and y = log x, then 



du 

dy 



i+y' 



dx 



and 



du 
dx 



s(i + y % Y 



6?# 



If w — i^(v, y), v =f(x) and ?/ =/,(#), to obtain 
total derivative of w with respect to x, we may proceed thus 

du , 



, the 



Since 



dividing by c?a?, 



dx 



dv dy *' 

du dv du dy 
dv dx dy dx' 



(2) 



du . 



which gives y 1 - in terms of derivatives which can be reckoned 
cix 

out from the given equations. 

Thus, if u = v* -\- vy, v — log x and y = e x , then 



du _ du dv 1 , 

— = 2v -f- y, —- — v, -T- = — and 
^ J dy dx x 



dy 
dx 



= e x ; 



72 DIFFERENTIAL AND INTEGRAL CALCULUS. 

substituting in (2), we have 

du 2v + y 

~r = L -^- + ve x . 

ax x 

If u = F(x, v, z), v —f{x) and z =f l (x), we have 

r 7 -, du _ du , , du , 

rWwl _ e?w f?w dv du dz 

\_dx J ~~ dx dv dx dz dx 9 ^ ' 

are partial derivatives, and — the 



, du du .. du 

where -7—, -r- and -7- 

cte ow dz 

total derivative of w. 

EXAMPLES. 
Find -j— in the following: 

1. u — e x (y — z), y = sin x, and z = cos x. 

2. « = tan"' |, and y = ^^^i 5 ! [g] = - -^ 



3. u = tan * (#y), and ?/ = e x . 



Vr 2 - a; 2 * 



dxj l + zV*' 

4. If # = uz and w = e x , z — x* — 4z 3 -f 12z 2 — 24a; -f 24, 

find the slope of the curve of which y is the ordinate and x the 

abscissa. dy „ . 

— --u — e x x . 
dx 

106. Successive Partial Differentials and Derivatives. 
We have seen that the differential of u (= f{x, y) ) (1) with re- 
spect to x is denoted by -r-(u)dx, and (2) with respect to y by 



GENERAL DIFFERENTIATION. 73 



Similarty, the differential of —dx 



d [ du , \ , d 2 u dx* 



(1) with respect to x is denoted by "i—l—r-dx \dx, 



, rt v " -. , . t , t , d ( du .. \ _ rZ 2 « rfo *?■?/ 

(2) with respect to ?/ is denoted by —\-j—ax\dy ~ 



dx\dx J ' dx~ 

d ( du \ _ d 2 u d 

dy \dx ' I J ' dx dy 

d*n du dx 
Hence — - -^ — is a symbol for the result obtained by dif- 
a y ax 

ferentiating u two times in succession : once, and first, with 

respect to y, and then once with respect to x. Similarly, 

- — -z—odx dip indicates the result of three successive differentials 
dx dif J 

of u\ first, once with respect to x, and then twice with respect 

to y. 

In finding these successive partial differentials, we treat dy 
and dx as constants, since y and x are regarded as independent 
variables, see Art. 68. 

The symbols for the partial derivatives are 

<Fu d*u d*u d 3 u d 3 u 
ctx*' dx~dy' dtf' dtf' dx^' 

d 2 u d*u 



107. Principle. If u =f(x, y), 



dxdy dy dx 

d*u 
For, Art. 102, changing x and then y to obtain , , or 

d 2 u 
changing y and then x to obtain , is equivalent to changing 

ay ax 

x and y simultaneously; and therefore the results are equal. 

Cor. I. If u be differentiated m times with respect to x, and 

n times with respect to y, the result is the same whatever be 

the order of the differentiations. 

EXAMPLES. 

1. Given u = a; 2 ?/ 3 ; find ^ — =- and -= — 7-. 6xy 2 . 

J dxdy dy dx a 



74 DIFFERENTIAL AND INTEGRAL CALCULUS. 

2. Given u = x>y + xf; verify gjfr = J^-. 

3. If u = y log (1 -f xy), show that 



dy dx dx dy 

„. x* — y 2 .. ^ 2 w d 2 u 

4. G-iven w = 2 , , - : verify - — —- = - — - . 
# + V dx dy dy dx 

108. To find the successive differentials of a function of 
two independent variables. 

Let u = f{x, y); then 

&$=%**+%** a) 

dit du 

Differentiating (1) and observing that j— and -j~ are usually 

functions of x and y, and that x and y are independent, Art. 68, 
we get 

r rfV J = w dx * + Sk dx dy + wk dy dx + w df ' 

or, Art. 107, 

Differentiating (2), remembering that each term is a function 
of x and y, we have 

[VTwl — ~--dx* 4- 3 7 „ ., dx* dy + 3^ — ^-jdx dy 1 + -r~rdy 9 . . . ; 
L J ^ 3 ' efo 2 d2/ J ' dxdif * ' d*/ 3 * 

and similarly may [d\i\, [d*u], etc., be found. By observing 
the analogy between the values of [d 2 u] and [d*u], and the de- 
velopments of (a -f- #) 2 and (a -f- ^) 3 , the formula for the value 
of [rf"«] may be easily written out. 

109. Implicit Functions. In functions of the form 

dii> dit 

f(x, y) — 0, the formula [du] = j-dx + -r— dy is often useful in 

dv 
finding the value of the derivative, or slope, -ys 



GENERAL DIFFERENTIATION. 75 

Thus, take f(x, y) = ax* -f- x sin y = 0. 
Making u = ax 3 -j- x sm y> we nave 

\du~\ = -j—dx -f- -7- dy = (Sax 2 -j- sin y)dx -{- x cos y dy. 

But since u = 0, [du] = 0. 

du 
dy _ dx 3ax* -f- sin y 

dx ~~ du ~ x cos y 

dy 

EXAMPLES. 

du 
Find the derivative, ~, of the following: 

-. / zn2 3 . 2 dy 3X* — 2ax 

2. *• + 3d*y - «y» = 0. J.' = f + 4 f * . 

3.^+3^+^=0, fe = - *;+ < y . 

1 ^ ' y dx y* -\- ax 

110. Successive Derivatives of an Implicit Function. 

The following examples will serve to illustrate how the succes- 
sive derivatives of implicit functions in general may be deter- 
mined. 

EXAMPLES. 

d*y 
1. Find ~~ Q when y 2 — kax = 0. 
ax 

XT dy 2a 

Here j = — (1) 

ax y v ' 

Differentiating (1), we have 

d*y _ - 2ady ( 

dx ~ y* w 



76 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Eliminating dy in (1) and (2), we have 

^l - _ 1?! 

dx 2 ~ if 

2. Given a'lf + ftV - « 2 6 2 = 0, show that %%r «= - -jL 

3. Given y s + z s - 3ff»y = 0, show that ^f= - -^BL. 

111. Change of the Independent Variable. After ob- 
taining the derivatives ~~, ^~, ~, etc., on the hypothesis that 

x was the independent variable and y the function, it is some- 
times desirable to change these expressions into their equivalents 
with y for the independent variable and x the function, or with 
x and y for the functions and some other variable, as t, for the 
independent variable, and so on. 

dti 

112. To find the successive derivatives of ~ when 

ax 

neither x nor y is independent. 

Under this hypothesis -j- is to be differentiated as a fraction 

having both terms variable. 

d?y _ d (dy\ _ dx d*y — dyd*x ( . 



dx* dx\dxj dx 3 

cr -i i d 3 y d (cFy 

Similarly, ^ = &(^ 

_ (*fo ^? 3 ?/ — ^?/ d 3 x)dx — 3(dx d 2 y — dy d 2 x)d" i x . 

~ : ! dx 1 ! ' ' * W 

In like manner we obtain the other successive derivatives. 
Cor. I. If y is independent, then d 2 y = d 3 y = 0, and we have 

d*y dy d*x 



dx 2 ~ dx 3 ' ' ' ' ' 

d*y _ 3(d 2 xydy - d*x dy dx 
dx 3 ~ " dx b 



(3) 
(4) 



GENERAL DIFFERENTIATION. 11 

Formulas (1) and (2) give us the values to be substituted for 

-t4 and y^ when neither x nor y is independent; and formulas 

(3) and (4) give us the values of the same derivatives when y is 
independent. 

If a new variable t, of which x =f(t), is to be the inde- 
pendent variable, in Art. Ill, we replace x, dx, d 2 x, etc., by 
their values as determined from x =f(t). 

EXAMPLES. 

1. Given yd 2 y -f- dy 2 + dx 2 == 0, where x is independent, to 
find (1) the transformed equation in which neither x nor y is 
independent; also (2) the one in which y is independent. 

d 2 v 
(1) Dividing by dx*, substituting for -~ f rom 0)> an d multi- 
plying both members by dx 3 , we^have 

y(d 2 y dx — d 2 x dy) + dy 2 dx -\- dx 3 = 0. 
(2) Making d 2 y = in this last equation, and dividing by 
— dy 3 , we have 

d 2 x dx s dx _ 
y dif dif dy ~ 

2. Change the independent variable from x to t in 

** + \*!L + y = o, when x = %Vl 
dx xdx 

d 2 y 
Substituting for -~ from (1), multiplying by x dx 3 , and mak- 
ing x = 2VI, dx — t~ i dt, and d 2 x = — \t~^df, we obtain 
d 2 y dy 

Change the independent variable from x to y in the two fol- 
lowing equations: 

Wary c?» efc 3 dx 2 \dx) ' dy 3 dy 2 

&y dif 3 d\f _ d 2 z dfz 2 _ _ 

4 * *^ 2 + dx~ 3 ~ tic ~ °' X df + dy 2 l ~ U * 



78 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Change the independent variable from x to t in the two fol- 
lowing equations : 

K d 2 y , 2x dy , y _ , 

5 - a£ + r+p I* + (i+iy = °' where • = tan *• 

6 - < x - ^S - 41 = °' where * = cos *• S= °- 

7. Find the value of R = f 1 + T^j ■*■ ~r^, where x is in- 
dependent, supposing neither x nor y to be independent. 

R = J&f+Jgf^ 

dr ^ 2 «/ — dy *f# 

MISCELLANEOUS EXAMPLES. 
Differentiate the following: 

1. y = 3(as 2 + l) § (4as 2 - 3). dy = 56x\x 2 + l)*tfas. 

1 dy x 

2. y = ■ -=. -f- = . - 1. 

x H- Vl + a; 2 «& Vl + a 1 



_ Va + x + ^a — a; dy a*-\-aVd *—x* 



4. y = x-\- log cos f — — as). 



7r \ dy 



9 



dx 1 -f- tan a; 

Va + |/ie dy_ Va 

5 ' y - l0g VTZ~^ dx~ {a - x) vV 

x log X , _ #i . dy lo & ^ 

„ , A M - x + 1 dy__ g - 1 

7. y - log y jq- x + r dx - ^ + x , + r 

8. y = sin (as + a) cos (x — a). dy = cos 2a;das. 

9. y = log tan (^ + - J. ^ = sec as. 

10. sin 2a: = 2 sin as cos x. cos 2as = cos 3 x — sin 2 x. 



GENERAL DIFFERENTIATION. 79 

. n 2 tan x 1 — tan 2 x 

11. sm 2z = .,,,„ . cos 2x — — — - — j— . 

1 -f- tan a; 1 -j- tan x 

12. sin 3a; = 3 sin a; — 4 sin 3 a?. cos 3x = 4 cOs 3 a; — 3 cos a;. 



■1 /,CC 



% 1 



13. y = tan 
^ aa; e* + e - * 

^-lX dy 6x* 

14. ?/ = COS M 6 , -, . -T~— « T-T. 

^ Va 6 -f 17 dx x e + 1 

. 1 dy 2 

lO. 1/ = SCO -1 rr-z r. ~-= . 

J 2a; 2 — 1 dx tfl — a. 

16. y = tan" 1 a' + tan" 1 ^-^. ^- = 0. 

. 1 -j- a; da; 



Prove the following by differentiation: 

17. / — = r — a— ^^ — = tan -1 in tan a;) + C. 

J cos a) -j- n sin a: v y 

1 f*2ax*dx x 1 _ /as — a\ 

is - y f^t« = tan « +2 lo ^ &+i/ + « 

19. lo g (l + z)==*-f+|-f + etc. 

20. tan -1 x = x — '-- + — — — -f- etc. 

o 5 7 



Find the slopes of the following curves: 
21. The quadratrix, y = (a — x) tan — . 



dy ?r / . 9 nx , nx 

-/- = ■— (a — x) sec 2 tan — -. 

dx 2a 2a 2a 



y 



22. The c} T cloid, x = r vers -1 \/2ry — y 1 



dy_ _ J2r - y 



80 DIFFERENTIAL AND INTEGRAL CALCULUS. 

c i - —~\ 

23. The catenary, y = -\6 >c -f e 7. 

(V 

24. The tractrix, x = a log 



-M I- 


d\j If* -*\ 


y 


?j 2 


£- - f fl a - y\ 

dy _ y 


6 

x 2 


dx y a * _ y * 



Find the following: 

^ 3(x 2 + 2) 3 
27. /*(§«* - 5^ § + 2af* + ix~ % )dx. (X ~ 1)3 + G 



/"iV- ^ ( a » _ x *y 



17 z 4 3aV ' 

29. f—— 3 . —J!L==+a 

J (a? -xy (fVd 1 -x* 
t ' ^ x V2ax -"?" a« 

01 /V ~ 2 )^ 2 6 , *" A 1 LAV 

33 '/K3^- log(, 2 + 2, + 3) + a 

33. jT^=ll& - 2 log (2s + 3) + 0. 

n t r dx , . /Sic — 2\ , ^ 

35. /"— £L=. $ log (2x + f/f+4?) + a 

36./., **,-„■ log(>-24-*V-4z+13)4-C. 



GENERAL DIFFERENTIATION. 81 

__ P dx . _ x 2x - 3, 

38. / sin -1 — -=- + (7. 

^ Vl + 3x- x* Vl3 

39. / -i — -— . -^ tan" 1 =r -f C. 

«/ x 2 — 6x -f 11 f ^ 4/ a 

40. / (z 2 -2z+2)(£-l)d£. -f. 


/oXCtX a , — — — . 

v V125. 
2tV -4 
2 

/^dx 



43. / 3 sin 2 x cos xdx. sin 3 x -f- C. 

44. / 36(a — & cos 2 <&)* sin jc cos 2^. (a — b cos 2 a;)*-f (7. 

45. /- 4— dx. log (tanse + aO+C. 

«/ tan a; -f a & v 

46. / (tan x -f- cot xfdx. tan # — cot x -{- 6". 



47. Find f| when y 2 - 2xy + c = 0. y^ ^. 

«# (y x) 

48. Given y 2 — 2axy -f a; 2 = c, to find —. - ^. 

ttiC lx^ (XX j 

49. In #^ 2 y + ^ 2 + d x * — 0; change the independent vari- 
able from x to y. 6f x ^a; 3 dx __ 

dy 2 ~ dy 3 ~ dy ~ 

50. Change the independent variable from x to z in 

(2a - l)«g + (2* - l)g = By, where to =1 + •. 

51. If «/ 2 = sec 2x, prove that y ~ = 3# 5 — ?/. 



82 DIFFERENTIAL AND INTEGRAL CALCULUS. 

52. Given s = f — hf + 6t; find the velocity (v) and its 
rate of change (a') when t = 10. v = 206; «' = 50. 

53. In t seconds after a body leaves a certain point the rate 
of change of its velocity is Qt — 12 feet per second; required its 
velocity and distance from the point of starting. 

v = 3f - 12*; s = f- 6t\ 

54. In the last example how far will the body travel in 10 
seconds ? _ ^ + [s f« = 32 + 432 = 464 (ft>)# 

4 

55. How many times faster is x increasing than log x, when 
x = ?i? u times. 

56. Required the value of x at the point where the slope of 
the curve y — tan x is 2. 7t 

T' 

57. A man is walking on a straight path at the rate of 5 ft. 
per second; how fast is he approaching a point 120 ft. from the 
path in a perpendicular, when he is 50 ft. from the foot of the 
perpendicular ? l^f ft. per sec. 

58. A vertical wheel whose circumference is 20 ft. makes 5 
revolutions a second about a fixed axis. How fast is a point 
in its circumference moving horizontally, when it is 30° from 
either extremity of the horizontal diameter ? 50 ft. per sec. 

59. A buggy wheel whose radius is r rolls along a horizontal 

path with a velocity v' ; required the velocity (-=7-) of any point 
(x, y) in its circumference; also the velocity of the point hori- 
zontally {jij and vertically (-77). 

The curve described by the point in the circumference of the 



y 

wheel is- a cycloid whose equation is x = r vers -1 - — \try — if\ 
differentiating this and dividing by dt, we have 

dx _ y dy 

dt Y2ry - y' 2 dt K) 



GENERAL DIFFERENTIATION. 83 



Again, the abscissa of the center of the wheel is r vers" 1 —\ 
differentiating this and dividing by dt, and we have 

,' = — ^ f (*) 

V2ry -ifdt 



Again, since ds 2 = dx* -f- dy 2 , we have 



ds_ _ J (dxV (dy_ 
dt~ y \dtl ~r\dt 



1 + [± (3) 



From (1), (2), and (3) we readily obtain 

60. In the last example find the values of -^— , —-, and ^- at 

r dt dt dt 

the point (1) where y = 0; (2) where y = r; (3) where ?/ = 2r. 

61. Water is poured at a uniform rate into a conical glass 3 
inches in height, filling the glass in 8 seconds. At what rate is 
the surface rising (1) at the end of 1 second? (2) At what rate 
when the surface reaches the brim ? 

(1) \ in. per sec. (2) -J in. per sec. 



CHAPTER V. 

SERIES, DEVELOPMENT OF FUNCTIONS, AND 
INDETERMINATE FORMS. 

SERIES. 

113. A Series is a succession of terms following one another 
according to some fixed law. 

If the sum of the first n terms of an infinite series approaches 
a definite limit as n increases indefinitely, the series is Conver- 
gent ; if not, it is Divergent. 

The sum of a finite series is the sum of all its terms; and the 
sum of an infinite convergent series is the limit which the sum 
of the first n terms approaches as n increases. An infinite 
divergent series has no definite sum. 

114. To Develop a function is to find a series, the sum of 
which shall be equal to the function. Hence the development 
of a function is either a finite or an infinite convergent series. 

For example, (x + l) 3 = x* + 3.i' 2 + 3x + 1. 

This finite series is the development of the function (x + l) s 
for any value of x. 

Again, by division we obtain 

= l-fa; + ^-f^ + ... x n '\ ... (1) 



1 -x 

Now this series is the development of only for values 

84 



SERIES. 85 

of x numerically less than 1, for the omitted remainder is 



x h 



1-x 



; hence, denoting the series by s, we have 



1 x r ' 



l-*~* + l-af 

1 x n 

Therefore s can be the value of only when = 0, 

1 — x J 1 —x ' 

and s the development of only for such values of x as will 

x — X 

x n 
cause to approach 0, as n increases indefinitely, and this 

_L — X 

can be the case only when x < 1. 
Thus : (1) For x = 2 we have 

-1 = 1 + 2 + 4 + 8 + ... 2"- 1 - 2% 

which would be absurd were the remainder — 2 n omitted. 
(2) For x = i we have 

2-i + I + L , I . J_ + l 

in which the remainder — decreases as w increases, indefinitely. 

115. A series is said to be absolutely convergent when it 
remains convergent on making the signs of all its terms positive; 
but only conditionally convergent when it becomes divergent on 
such a change of signs. 

The series 1 — i + 3 — i + . . . is an example of a condi- 
tionally convergent series. 

1 16. Prop. TJie infinite series u x + w 2 + . . . u n _ x + u n + . . . 

will be absolutely convergent if the terms are all finite, and the limit 

11 
of the ratio — - , as n is indefinitely increased, is 7iumerically less 

than unity. 



86 DIFFERENTIAL AND INTEGRAL CALCULUS. 



u 
Since the limit of — - -, as n is indefinitely increased, is less than 

1, there must exist some finite integer, c, such that for all values 

u 
of n which are greater than c, — — is less than 1. 

The sum of the terms u x + ? ' 2 ~\~ • • • u c is a definite finite 
quantity, and it only remains to show that the sum of the remain- 
ing terms, u c+1 -\- u c+2 +.-...., is also. 

Let 7 be a number less than 1 but greater than any of the 

ratios 5**, %&, ^±2, . . . ; then 
u c u c+l u c+2 

U c ^i <C T 1l c , 

u e+a < ru c+1 < r*u c , 

u c+s < r u c+2 < r* u c+1 < r 3 u c9 



or 



Mo+i + Uc+2 + u c +z + . . . < u c (l -f- r + r 2 + . . .)r, 
u cH + u c+2 + u c , z + . . . < ruj^— -J, 



the last member of which is finite, since r < 1, Art. 114; there- 
fore the first member is also finite. 

Again, that the sum u c+1 -f u c+2 -f . . . , though finite, may be 
indeterminate, is precluded by the fact that the limit of u n) as n 
approaches oo , is 0. Therefore the series is convergent. 

Scholium. The limit of u n as n increases is always in case 
of a convergent series, but the mere fact that the limit of its 
nth. term is does not prove a series to be convergent. 

117. Cor. I. The series a + a x x + ajf + . . . a n _ x x n ~ x 
-\- a n x n -f- . . . , where a , a lf a 9 , etc., are independent of x, is con- 
vergent for all values of x numerically less than k (say), the limit 

of - s= ^, as n approaches oo . 

For, by Art. 116, the series is convergent if the limit of 



SERIES. 87 

a n x n -7- tf„-i£ n_1 , or a n x -±- a n _ l9 is less than 1, and therefore if 
x < k. 

118. Cor. II. When x > k the series is divergent, and when 
x = k the series in some cases is convergent, and in others di- 
vergent. 

119. Cor. III. The series f'{x) = a, + 2a 2 x + 3fy? 2 -f . . . 
(n — l)a n _ x x n ~ % -f- ?ia n x n - 1 , obtained by differentiating /(») = a 
-f- «,£ -f ft 2 ^ 2 H~ • • • a n-iZ n ~ 1J r a n % n , is convergent for the same 
values of x as the last-mentioned series. 

For in the former k — the limit of '--^. which is the 

na n 

same as the limit of — — . 
a n 

It is also evident that the limits of convergence of the series 

obtained by integrating the individual terms of f(x)dx are the 

same as those of the series f(x) itself. 

EXAMPLES. 

Find the values of x which will render the following con- 
vergent : 

x 9 x a x"- 1 x n , 

Here a n _ x = and a n = -. .*. — = = 1 A , 

n — l n a n n — 1 n— 1 

which = 1 when n = co . Hence (117), — 1 < x < 1; that is, 

x lies between — 1 and -f- 1. 

„ , , , z 2 , a; 3 , x* X"' 1 x n , 

Here -^ = - — =—— = n. which = co when n = oo ; hence 

a n \n — 1 ' 

-co < a; <co ; that is, the series is convergent for all finite 
values of x. 

3 ?? , 52! ■ ?* , , 2(«-l)+l , 2» + l 



88 DIFFERENTIAL AND INTEGRAL CALCULUS. 

jr ff»-i" 2(n--l)'+l ft 2 + l ,:■ : •, 

Here = ~ ' x = — , which = 1 when n =00 . 

a„ (» — 1) + 1 2n + 1 

.'. - 1 < x < 1. 

4. z + 2 V -f 3 V + ...(n- 1) V- 1 + »V -]-... 

- l<z < 1. 

5, g + *', + *"l_ + _-g. = g. 
1-f^r 1 + ^ " "l + ^w-l l + Vn ' 

- 1< a; < 1. 

DEVELOPMENT OF FUNCTIONS. 

120. There are two common and useful formulas for de- 
veloping functions : Taylor's and Maclaurin's. 

We shall deduce Taylor's formula first, as Maclaurin's may 
be derived from it. 

121. Taylor's Formula is a formula for developing/^ + x ) 
in a series, where f(y -f- x) represents any function of the sum of 
two variables, such as (y + x) n , log (y + x), sin (y -f #), a y+x . 

The derivation of Taylor's formula may be regarded simply 
as the process of finding the acceleration of a function. Thus, 
let u = /(?/), and suppose y to be increased by x, we then have 
(Art. 24), ' ' 

4* = f(v + x ) - Ay) = Ax + m * x *> • • W 

or f(y + x ) = f( l j)J r Ax^m,x% (2) 

where A (= m l -=f f (y) ) is a constant with respect to x, and 
mj? is the acceleration of u (Art. 25), and this is what we wish 
now to determine. 

Since m n is of such a character that m^x vanishes with %, we 
will assume (see footnote, p. 10) 

m 2 = B + Cx + Dx* + . . . Lx n ~ 3 , .... (3) 

where B, C, D, . . . L are independent of x, and x has such a 
value as to render the series (if infinite) convergent. Substitut- 
ing in (2), we have 

f(y + x)=f(y) + Az + Bx*+Cx> + ...Lx n -\ . (4) 



DEVELOPMENT OF FUNCTIONS. 89 

Differentiating (-1) successively with respect to x, we have 

f'{y + x) = A + 2Bx-\-3Cx' + ...(n-l)Lx n - 2 ; (5) 

f(y + x) =2B + 6Cx + ...(?i-l){n-2)Lx n - 3 ; (6) 

f'iy + X ) =QC + ...(n- \){n - 2){n - 3)Zz n " 4 ; (7) 

• • J 

f n - 1 (y + x) = \n-lL (8) 

These equations, (5), (6), (7), etc., are true for any value of 
x which renders equation (4) convergent (Art. 119) ; therefore 
they are true when x = 0, which gives 

f'(y) = A, .-. A=f'(y); 

f"(y) =22?, .:B = ^f"(y); 

f'"(y)=eO, ... C=i-/'"G/); 



f-\y) = \ n-l L, ,-. L = -—j-/-^). 

Substituting these values for A, B, C, ... L in (4), we have 



(A) 



This is the formula required, which was first published in 
1715 by Dr. Brook Taylor, from whom it takes its name. 

The preceding is not a rigorous demonstration of Taylor's 
formula, inasmuch as the possibility of development in the 
proposed form is assumed. A rigorous proof, including the 
form of the remainder, has been inserted in the Appendix, A. , 
to be used or not, according as the teacher or student may 
desire. 



90 DIFFERENTIAL AND INTEGRAL CALCULUS. 

122. Cor. I. To determine for what values of a; the series 
is convergent. 

The nt\\ and (n -f l)th terms of the series are evidently 

/"W^T and /%)£. 



Therefore (Art. 117) the series is convergent for any value 
of x numerically less than 

jpvi + fM, or JfTM), when „ =00; 

\ n-i \n \ r\y) r 

f n ~l(y) 

Hence, if • . , / is not zero when n =oo , the series is con- 

f{y) f~(y) 

vergent for all finite values of x; and if n ~-j^ ( ^r — 0, when 

n = oo , the series is divergent for all values of x except 0. 

In deducing Taylor's formula we have supposed all the func- 
tions that occur to be continuous. Hence the formula is inap- 
plicable, or "fails," if the function, or any of its differential 
coefficients, be infinite for values of the variable lying between 
the limits for which the development holds. 

123. To develop (y -f x) m . 
Here f(y + x) = (y + z)~ 

Make x = 0, f(y) = y m . 

Differentiate, etc., f'(y) — my m ~ 1 , 

/"(y) = m(m - \)y m ~\ 
f"\y) — m(m - l)(m - 2)y m ~\ 
etc. etc. 

Substituting these values in (A), we have 

(y -f x) m = y m -f mxy m ~ l -f v |9 f x*y m ~* 

, m(m — l)(w? — 2) 3 m . , ,_. 

+ — r^~ L x 3 y m -\ etc., (B) 

which is the Binomial Theorem. 



DEVELOPMENT OF FUNCTIONS. 91 

Cor. I. Let us determine for what values of x the equation 
is true,, supposing m negative or fractional. 

f n ~ l (y) = m(m — 1) . . . (m — n + 2)tj m - n+1 , 

f\y) = m(m — 1) . . . (m — n + l)y m - n ; 

... (Art. 122), n^ {y) - 



f\y) m -n + ¥ 

which = — y when n = go . 

Therefore formula (B) is true when x is numerically less 
than y. 

Cor. II. Making y in (B) equal to 1, we have 

(l+x) m =l+mx+ — ^ V+-* ^ V + etc., (C) 

in which — 1 < x < + 1- 

124. To develop sin {y -\- x). 

Here f(y -\- x) = sin (y -{- x). 

Making x = 0, and differentiating, we have 

f{y) = siny; /'(#)= cos y; f"(y) = - sin y; 
f"\y) = - cos y ; f iv (y) = sin y ; etc. 
Substituting these values in (A), we have 

sin (y + a) = sin y (l - -|- + ~- -— |- + etc.\ 

L y f (D) 



/ ar . ■ ar z 7 , , \ 

+ cos ^f-|3"+|5"-r+ etc - 



Cor. I. In (D) by making y = 0, remembering that sin = 
and cos = 1, we have 

/y 3 /£ 5 /v T 

sin x = x _—+-—_ — -j- etc. ...(E) 



92 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Coe. II. Differentiating (E), we have 

x* x i x* 
a »s = l__+___ + cta. . . . (F) 

Cor. III. For the quantities within the parentheses in (D), 
substituting their values from (E) and (F), we have 

sin (y -f- x) = sin y cos a; -j- cos y sin #. . . . (G) 

Coe. IV. Differentiating (G), regarding a; as constant and «/ 
as variable, we have 

cos (y -{- x) = cos y cos # — sin y sin z. . . . (H) 

125. To develop log (y -\- x). 

f(y -\- x) = log (y -\-x); making x = 0, and differentiating, 
we have 

M = 1^ (y)> f'(y) = J-; /'%) = - £; 
/"%) = | s ; /'%) = - |r, etc. 

Substituting in (A), we have 

log (y + *) = log (y) + ^ - ^5 + Jp - etc., . . (I) 

which is the logarithmic series. 

Coe. I. The nth. and (n + l)th terms of (I) are, omitting 

xi • x *~ l ^ x " i a j. Vim a " _1 w y 

the signs, -. . - , , and — - ; hence, Art. 117, = — ^-, 

to (n — l)y l ny n a n n — 1 

which = y when n == go . Therefore formula (I) is true when 

# is numerically less than ?/. 

Coe. II. In (I), by making y = 1, we have 

Iog(i + *)=«-£ + y-5 + «to.. ■ • (K) 

which is true for all values of x numerically less than 1. 



DEVELOPMENT OF FUNCTIONS. 93 

126. Maclaurin's Formula is a formula for developing a 
function of a single variable, as y = a x , y — log (1 -\- x), 

y = (« + z) n . 

It may be derived from (A) by making y = 0, which gives 
f(x) = f(0) +f(0)x +/"(0) * + /"'(0)4 

+ --/- 1 (0) 1 ^ I + ..., (L) 



in which /(0),/'(0),/'"(0), etc., represent the values which /(a:) 
and its successive derivatives assume when x = 0. 

Coe. I: Formula (L) is true for all values of x numerically 

less than ^ , when w = x . 

127. To develop a x . 

Here /(z) = a x , .-. /(0) = a = 1; 

/'(&)= a* log a, .\/'(0) =loga; 

f"(x)=a*log'a, .-. f'(0) =log 2 a; 

/'"(*) = «* log 3 a, .-./'"(()) = log 3 a; 
etc. etc. 

Substituting these values in (L), we have 



a* = 



1 + log ax + log 2 a— + log 3 a-^r + log 4 a-^-, . (M) 



which is called the Exponential Series. 

Cor. I. This series is convergent for all finite values of x, 
since /" _1 (0) -^/ n (0) is obviously finite and different from zero 
for all values of n. 

Coe. II. Making a = e, remembering that log e = 1. we 
have 

^l + s + ^ + i^ + JjL + etc.. . . (N) 



94 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Cob. III. Making x = 1, we have 

Hence e = 2.718281 +. 

128. Find the development of tan -1 x. 

In the applications of Maclaurin's formula the labor of find- 
ing the successive derivatives can often be lessened by taking 
the development of the first derivative, as follows: 

/0) = tan" 1 ^ /./(0) = tan- 1 = 0: 

f ' {x) = Ih^ 2 = * ~ ^ + x * " x * + etc " ■'" /r(0) = * 

f'(x) = - 2x + 4x* - 6x b + etc., .\ /"(0) = 0: 

f"{x) = -2 + 3. 4. x 2 - 5.6z 4 + etc., -•./'"(O) = -2; 

/%) = 2 . 3 . 4:X - 4 . 5 . 6x 3 + etc., .*. f y (0) = °l 

f(x) = |4 - 3 . 4 . 5 . 6x 2 + etc., .'. / v (0) = |4 ; 

f*(x) = - 2 . 3 . 4 . 5 . 6£ + etc., ,\ /^(O) = 0: 

/^(a.) = _ |6_ + etc., .-. fi*<p) = - |6; 
etc., etc. 

Substituting in (L), we have 

tan- 1 ^ = x— — + y — y + etc. . , . (Q) 

EXAMPLES. 
Develop the following : 

x i . x 6 bx e 



1. 4/1 + ,'. 1+i .- ¥ + --_ + etc. 

Put # 2 = ?/, and develop; then replace y by its value. 

2. («-fa;)- 3 . a' z — 3a- 4 z + 6« V - 10a~ 6 x 9 + etc. 

3. (! + ,)»• i + 1-l+g.-ete. 



DEVELOPMENT OF FUNCTIONS. 95 



, , , tf X* x° x° 



K •.cos a; 



/, x' , Ax' Z\x° , . \ 



6. tan a. a; 4- — -4- — - + etc. 

o 10 

X 2 DX* 

7. sec a?. 1 + y -f — -f etc. 

2 3 4 

8. log (1 + sin *). x - |- + j - ^ -f etc. 

9. log(l + e*). log 2 + |+^ -etc. 



10. e* sin *. 1 + x* + ^ -f etc. 

2# 3 

11. e* sec ic. 1 -f a; -f a; 2 -f -— - -f etc. 

o 

12. log (1 - z + z ! ). -s + f' + ^+J-etc. 



Put — x -{- x 2 = y, and develop; then replace y by its value. 
In the two following examples put y for x 2 , develop, and re- 
place y by its value. 

i3 -' ra ! -■?■-?-■£ -&-•*■ 



1 . x 2 . 3a; 4 3.5z f 

Vl + x 2 



U ' ,/rr^i - 1 -2 + r4"IU + ek 



129. To find the value of nr. 

We find by development that 

, x 3 , 1.3a; 5 , 1.3.5a: 7 , 
sm -i fl; = ^ + _ + __ 5 + __ +e tc., 

where a; lies between — 1 and -J- 1. 



96 DIFFERENTIAL AND INTEGRAL CALCULUS. 



Making x = $, remembering that sin -1 -J = — , we have 

* J. I 1 _3_ 5 \ 

6 2 V + 24 + 640 + 7168 + J' 

or 7t = 3.141592 + . . . 

130. To compute Natural logarithms. 



We have found 

log (1 + 4= 3-^"+^- !- -etc.,. . (1) 



x 3 X* 



where x is numerically less than 1. 

We now proceed to modify this series so that it shall be true 
and convergent for larger values of x. 

Substituting — x for x in (1), we have 



Jj Jb Ju %J0 



\og(l-x) = -x--------- etc. . (2) 

Subtracting (2) from (1), we have 
log(l + ^)-log(l-^) = 2^ + ^+^+| 7 + etc). . (3) 

In (3) make x = , where z may have any positive 

value ; then - — ! — = — ! — ■ 

1 — x z 

and log (1 + x) - log (1 — x) = log (z + 1) - log (2), 

we have log (2 + 1) = log (z) 



+ 2 



1 , 1 1 , 



(4) 



L22 + 1 ' 3(22 + 1) 3 ' 5(22 + 1) 
By this series we can compute the Natural logarithm of any 

number (2 + 1) when we know the logarithm of the number (z) 

less by unity. 

Making 2 = 1, remembering that log (1) = 0, we have 



log (2) 



1 , 1 1 ' . 1 , j. "1 

3+3^ + 5^3>+77^ + etC -J 



DEVELOPMENT OF FUNCTIONS. 97 

Taking six terms of this series, we have 
log 2 = .693147+ . . . 
Patting z = 2 in (4), we have 

log (3) = log 2 + 2(1 + 3JL + _L + _L + etc.) 

= 1.098612 + 

log (4) = 2 log 2 = 1.386294 +. 

Putting z = 4, we have 

log (5)=log4 + ^ + ^ + ^ + ^- 7 + etc.) 

= 1.609437 +. 
log 10 = log 2 + log 5 = 2.302585 -f . 

In this way we can compute the natural logarithms of all 
numbers. It is not necessary to use the formula in finding the 
logarithms of composite members, for they can be found by 
simply adding the logarithms of their factors. Thus log 15 = 
log 3+ log 5. 

131. To compute common logarithms. 

The modulus of the common system is m = log 10 e (Art. 
79). Hence 10™ = e, .'. log (10 m ) = log e, or m log 10 = 1. 

Let log 10 v = n and log v = n'; 

then 10 n = v and e n ' = v; 

10 n = e n/ , log (10 n ) = log e n ', or 
n log 10 = n\ .-. n = (n') .434294 +. 

Hence, to find the common logarithm of any number we 
multiply the natural logarithm of that number by the modulus 
of the common svstem. 



98 DIFFERENTIAL AND INTEGRAL CALCULUS. 



INDETERMINATE FORMS. 

132. In algebra — is called a symbol of indeter ruination, 
since any number whatever may assume this form. 

Thus, wXO = 0: divide both sides by and we have n — -. 

There are many fractions which assume the form of — in 

consequence of one and the same supposition, which makes both 
numerator and denominator = 0. Snch fractions are called 
Vanishing Fractions, and their values, which appear under the 

form of — , can generally be determined by the calculus. 

/£ 3 d" o 

Thus the fraction -= becomes — when x = a. 

x — a 

This form arises from the existence of a factor (x — a) com- 
mon to both numerator and denominator, which factor becomes 
under the particular supposition. Dividing both terms by 
this factor, we have 

x s — a 9 x* + ax 4- a* , . , 3a . 

, which = — when x = a. 



sf-a* x + a ' 2 

133. To evaluate a fraction that takes the form of - 

Let u and v be functions of x such that when x = a, u — 0, 
and v = 0. 

Let u and v be estimated from the point where their values 
are 0, that is, from where x = a; then when a: (= a) is increased 
by h we shall have, identically, 

u _ Au 
v ~ Av 



INDETERMINATE FORMS. 99 

As h approaches 0, or x approaches a, the limit of — is 

du 
equal to -=— (Art. 27) ; therefore 



u "1 _ ^w 



which is read, when x = a, — is equal to — . 

Applying this to the preceding example, we have 

x 3 - a s ~\ _ d(x* - ft 3 ) ~l _3aH _3« 
z* - a 2 ] a ~ d(x* - d 2 )_\ a ~2zJ a ~ 2' 

An easy deduction of the rule for reckoning forms like 

— is obtained by the use of Taylor's formula, as follows : 

Let <p(z) and f(x) be two functions of x such that f(x) = 

and <p(x) = 0, when x — a, then we shall have -^J = -. 

f{a) 

t, . , ,, (p(a) limit [~ d>(a 4- h)~~\ 

_ limit [- 0(g) + 0»ft + |0'»A 2 + . .."] _ <p'{a) _ 0(z) ~| 
~h = 0lf(a) +f'(a)h + tf'\*W + - - J '" /» ~ /(^)J ." 

Cor. I. If 0'(«) = and/'(«) = 0, we obviously have 

0"(q) _ 0(a?n . 
/"(«) ~ /(*) J. ' 



and if (f>"(a) = 0, and /"(a) — 0, we have 

; and so on. 



0"'(a) _ 0(g)" 



Hence, Rule. Substitute for the numerator and denom- 
inator, respectively, their first derivatives, or their second 
derivatives, and so on, till a fraction is obtained whose terms do 



100 DIFFERENTIAL AND INTEGRAL CALCULUS. 

not both become ivhen x = a ; the values thus found will be the 
true value of the vanishing fraction. 

EXAMPLES. 
Compute the following: 

x* — 16 1 8 



- 20j, 



x' + x - 20j; 9 

5x* — 8x + 3~l 2 



a- 



7z 2 -9z + 2J 1 5 

z 3 - 3z + 2 ~| 3 



a- 



x 3 -x* -x + IX 2 

■]• 

_Jo 



_, 1 — COS X , 

5. : . 0. 

sin # 

a - V~aF^-~x~*^ 



« Jo 



9. 
10. 



sm x 



12. 



-u- 



sing-g cos af] 



2a 



fi ^ — sin af| 

« 3 ' in' 

] 

- 1 log sin x ~| 1^ 

(^-2^.* 8 



1 

6 

sin 3x ~1 3 

a; — | sin 2£_J * ~ 2 

e x — e~ 



2 

sec 2 x — 2 tan aTl 1 



1 + cos 4z X 2 

4 



a — sm a? J 
(e*~l) 3 J/ 6" 



INDETERMINATE FORMS. 101 

There are other indeterminate forms besides -, such as — , 

oo 

oo x 0, co -co, 0°, go °, 1°°, which will be considered in succes- 
sion. 

134. To evaluate a fraction that takes the form of — . 

00 

Let u and v be functions of x such that u =co and v =oo 
when x = a; then for the same value of x, - = and - = 0. 



Hence 



1 

u v , 

— = — = -, when x = a. 

V 1 ? 



u 



Art. 133, - = — — - = -r=-, when a; = «. . . . (1) 
v /1\ v^w' v 7 



Dividing (1) by — , we obtain 

udv . w~| _ £?w"| 

Now (2) is derived from (1) by dividing by — ; hence, if — is 

finite, (2) is true for all finite values of — ; and if - = or oo , it 

v v 

may be shown that (2) is true in these cases also. 

it 
Suppose ——0 when x = a, and k a finite quantity, 

u . 7 u + lev 7 

then — \- k = = k. 

v v 

To this last fraction (2) evidently applies, hence 

u + kv du + kdv u , , du , , 

— = ~ , or - + k = -=- - + k\ 

v dv v dv 

,, , . u du , 

that is, — = -^— , when x = a. 

v dv 



102 DIFFERENTIAL AND INTEGRAL CALCULUS, 



11 v 

If — — co , then — = 0, and we have the preceding case. 

Therefore the form — is to be evaluated in the same way as 



the form -. 



EXAMPLES. 



Find the values of the following; 

i. * n 



2. 



d(x) 
~ d(\ogx) 

ax* -}- lr 

ex* -f- d_j 

3. ^=S\ . 

X J 

5. 
6. 



i=n =*i 






tan x 
log x~ 



1 



1- 



log tan 2x "1 
log tan x J*" 



log cot a; 
cosec x 



tan^-(z + l)j 



7TX 

tan — 



9. 



log cos (inx) 
log (1 - x) 



1 



a 

c ' 

CO. 



0. 

0. 
0. 

1. 



INDETERMINATE FORMS. 103 

135. To evaluate a function that takes the form of X °o 

or oo — oo . 

Functions of this kind can be transformed so as to assume 

the form of - or — , and then be evaluated by the previous 

methods. 

EXAMPLES. 

Find the following: 

5 
1. sec 3x cos bx] . — -. 



This takes the form of oo X 0; but sec 3rs cos bx 



cos bx 
cos 3x' 



which takes the form of . 
2. sec x — tan x] r . 0. 

This takes the form of co — oo ; but sec x — tan 



sma; 1 — sin x . 

= , which takes the form oi -. 

cos x cos x 

1 

log 



)gx a — lj, 



? X s . 



4. cosec 

e x — e x 
6. (1 — tan x) sec 2x]„ . 1. 



cos # 



„ x — a\ 7TX 

7. — tan — - 

a 2a 



8. x sin 
a; 



_ a- 

aT 



9. # m log n x\. (m and ft being -j-.) 0. 

2 
10. (1 - • x) tan (Itt^)],. 



7t 



104 DIFFERENTIAL AND INTEGRAL CALCULUS. 

136. To evaluate a function that takes the foim of 0°, co °, 

-I 90 

or 1 . 

Take the logarithm of the given function, which will as- 
sume the form of X co , and can be evaluated by Art. 135. 
From this the value of the function can be found. 

EXAMPLES. 

Find the following: 

This takes the form of l 00 . Making y = (l -J- - ) , we have 



log y = x log (1 +"—')• The value of x log f 1 -j — ) is found 
by Art. 135 to be a. Hence, when x = oo , log y = a; .'. y = e a , 

2. Vx]^, or x*J m . 1. 

1 1 
This takes the form of oo °; the log of x x is -logx, the value 



i 

is 

x 

s 5 ! = i. 



of which, when x = go , is 0; hence x a 

3. (sin z) tan % . 1. 

2 

4. (cot x) sin *] . 1. 

5- if + 1)3.. e. 

1 



6.(tanf)-?]. 
7. (cot a) 1 " 3 ^],. 



e 
1 

e ' 
i 

8. ftog (e + *)]„. e J 

9. tV + z] . x 
10. ^cos 2.r] . ?' 



INDETERMINATE FORMS. 105 

11. fS*] • i. 

X —I oo 

12. (cos^a^X e~ inm \ 

137. In implicit functions, as f{x, y) — 0. the derivative 
-— can be evaluated by the previous methods when it assumes 
an indeterminate form for particular values of x and y. 

EXAMPLES. 

1. Find the slope of x A — a*xy -\- b*y* at the point (0, 0). 

TT dy Ix' — cfy , . 

Here A = te=2 ¥y = WheD X = y = °- 

a?/ a# ax ^ 

-^- — ^— = Ti when a; = y = 0: 

that is, | = * or W®' -<(£)=*£ 

a 3 — 2^V 
dx 

dy . a 2 

,\ / = or ^. 

£?z £ 2 

2. Find the slopes of a 3 — 3azy + # 9 = at (0, 0). 

and co . 

3. Find the slopes of x* -f- ax*y — ay z = Q at (0, 0). 

and ± 1. 

4. Find the slopes of y 2 = x(x + a) 2 at (— a, 0). ± V — a. 

5. Find the slopes of x* -f 2«a; 2 ?/ — aif = at (0, 0). 

and ± 1/2. 



CHAPTER VI. 
MAXIMA AND MINIMA. 

DEFINITIONS AND PRINCIPLES. 

138. A Maximum Value of a function is a value that is 
greater than its immediately preceding and succeeding values, 
and a Minimum Value is one that is less than its immediately 
preceding and succeeding values. 

Thus, while x increases continuously, if f(x) increases up to 
a certain value, say/(«), and then decreases, f(a) is a maximum 
value of f(x) ; and if, while x increases, f(x) decreases to a 
certain value, say/(£), and then increases, f(b) is a minimum 
value off(x). 

For example, sin x increases as x increases till the latter 
reaches 90°, after which sin x decreases as x goes on increasing; 
that is, sin 90° is a maximum value of sin x. 

Again, if x increases continuously from to 6,f{x) = x* — 6x 
-f- 10 will decrease until x becomes 3 and then it will increase; 
hence /'(3) = I is a minimum value oi f(x), or x* — Qx + 10. 

Let the student substitute 1, 2, 3, . . . 10, successively, for x 
in /(a) = x 3 — \%x l + 96z - 20, and thus show that /(4) is a 
maximum and/(8) is a minimum. 

139. Any value of x that renders f(x) a maximum or a 
minimum is a root of the equation f {x) = or go , if f(x) and 
f'(x) vary continuously with x. 

For, if we conceive x as always increasing, f(x) changes from 
an increasing to a decreasing function as it passes through a 

106 



MAXIMA AND MINIMA. 107 

maximum value, say/(«), and from a decreasing to an increas- 
ing function as it passes through a minimum value, say f(b). 
Consequently /'(a) must change sign as x passes through a or b, 
Art. 25. But f'(x) can change sign only by passing through 
or oo . Therefore f'(a) or f'(b) = or go ; that is, a and b are 
roots* olf'(x) = or oo . 

To illustrate the preceding principles and definitions 
graphically, let y = f(x) be the equation of the curve Am ; then 
f'(x) = the slope of the curve at 
the point P or (x, y), Art. 48. As 
x (= OB) increases, the point P 
will move from A along the curve 
to the right, and y or f(x) will in- 
crease till it becomes aa', and then °' a B b c de h m ' 
decrease till it becomes bb', and 

then increase, etc. Therefore aa\, cc/, ee' are maximum, and bb', 
dd' are minimum, values oif(x). The slope of the curve, f'(x), 
is evidently positive before, and negative after, each maximum 
value of f(x); and negative before, and positive after, each 
minimum value of f(x). Moreover, at the points where f(x) is 
a maximum or a minimum, the curve is either parallel or per- 
pendicular to the axis of x, and therefore/' (x) = or qo , 

The converse of this theorem is not always true; that is, any 
root of f'(x) = or oo does not necessarily render f(x) a max- 
imum or a minimum. It is our purpose now to determine 
which of the roots will render f(x) a maximum and which a 
minimum. 

140. If the sign of f'{x) undergoes no change as f'(x) passes 
through or go , the corresponding value of f(x) will be neither 
a maximum nor a minimum. 

For, so long as the sign of f'(x) undergoes no change, f(x) 



* Here, and in what follows, the word root includes the real values of x 
which satisfy the equations f'{x) = or oo , whether/'^) be an algebraic or 
a transcendental function. 



108 DIFFERENTIAL AND INTEGRAL CALCULUS. 

does not change from an increasing to a decreasing function, 
nor vice versa. 

Cor. I. If an even number of the roots of f'{x) — or oo 
are equal to a, then x — a will not render f(x) a maximum or 
a minimum. 

141. If the sign off'{%) undergoes a change as f'(x) passes 
through or oo, the corresponding value of f(x) is a maximum 
or a minimum. 

For, if fix) undergoes a change of sign, f{x) necessarily 
passes from an increasing to a decreasing function, or vice versa. 

Cor. I. If an odd number of the roots of f'(x) = or oo are 
equal to a, then x = a will render f(x) a maximum or a minimum. 

Cor. II. Therefore, omitting the equal roots of which the 
number is even, every real root of f'{x) = or go will render 
f(x) either a maximum or a minimum. Now, of these let us 
find which will render f(x) a maximum and which a minimum. 

142. Maxima and minima of a function occur alternately. 
For, suppose that f(a) and f(b) are maxima of f(x), where 

a < b. Just after passing through f(a),f(x) is decreasing, and 
increasing just before it reaches f(o); but in passing from a 
decreasing to an increasing state it must pass through a mini- 
mum; hence there is one minimum between every two consecu- 
tive maxima. 

Cor. I. Denote the roots oif'(x) — and /'(#) = co which 
will render f(x) a maximum or a minimum by a lS a 2 , a 3 , a A , 
etc., in ascending order or algebraic magnitude. Then, if /(#) 
is an increasing function for all values of x less than a lf that is, 
if f'(a 1 — h) is positive, h being ever so small, /(a x ),/(rt 3 ), f(a b ), 
etc., are maxima, and f(a 2 ),f(a A ), etc., are minima; and if f(x) 
is a decreasing function for the same values of x, that is, if 
f'(a x — h) is negative, the maxima and minima will be inter- 
changed. 



MAXIMA AND MINIMA. 109 



RULES FOR FINDING MAXIMUM AND MINIMUM VALUES OF 

FUNCTIONS. 

143. The preceding principles indicate the following rule 
for finding the values of x which will render any function as 
f(x) a maximum or minimum: 

Differentiate the function f(x); make /'(%) = and 
f'(x) = qo ; find the real roots of both equations, and arrange 
all of them in order of algebraic magnitude, as a 1 , a. 2 , a 3 , etc., 
omitting the equal roots when there are an even number of 
them; substitute — oo or a l —h, li being very small, for x in 
f'(z), and (1) if the result is -J-, a i9 a 3 , etc., will each render 
f(x) a maximum, and a 2 , a A , etc., will each render f(x) a min- 
imum; (2) if the result is —,/(rt 1 ),/(rt 3 ), etc., will be minima, 
and /(ff a ), /(aj, etc., will be maxima. 

144. The preceding rule requires that all the real roots 
shall be found; it is sometimes desirable to know independently 
whether any particular root as a' will render f(x) a maximum 
or minimum. This may be done thus: 

I. Substitute a' — h and a' -\- h for x in f'{x), h being a 
small quantity, and (1) if f'(a' — h) is -f, and /'(a' -f- h) is — , 
f(a') will be a maximum; (2) if f f (a 9 — li) is — and f'{a! + h) 
is -\-,f(a r ) will be a minimum, and if f'(a' — h) and/'(&' -\-h) 
have the same sign, /(a') will be neither a maximum nor a min- 
imum. 

145. II. Developing f(x — h) and f(x + h) by Taylor's 
formula, substituting a' for x, transposing /(a'), and remember- 
ing that f{a') = 0, we have 

/i«' - *) -/(«') =/'>')! -rwj +/ Lv K)|r - (i) 

t V- li 

and 



110 DIFFERENTIAL AND INTEGRAL CALCULUS. 

If li be taken very small, the sign of the second member of 
either (1) or (2) will be the same as the sign of its first term. 
Hence, if /"(«') is negative, f(a') is greater than both f(a' — 7i) 
and f(a r -(- li), and therefore a maximum; while if /"(«') is 
positive, f(a') is less than both f{a' — li) and f{a' -f h), and 
therefore a minimum. If /"(«') = 0, and /'"(«') is not 0,f(a') 
is neither greater than both f(a f — h) and f(a' -\- h) nor less 
than both, and is therefore neither a maximum nor a minimum. 
If /'"(«') as well as f"(a') is 0, then, as before, f{a') will be a 
maximum or a minimum according as f"(a') is negative or po- 
sitive; and so on. 

Hence if a' is a root off'(x) = or ho, substitute it for x in 
the successive derivatives of f(x). If the first derivative that 
does not reduce to is of an odd order, f(a') is neither a maxi- 
mum nor a minimum; but if the first derivative that does not 
reduce to is of an even order, f{a') is a maximum or a mini- 
mum, according as this derivative is negative or positive. 

Note. — In many instances this rule is impracticable on ac- 
count of the great labor involved in finding the successive de- 
rivatives. 

146. The following principles are self-evident, and often 
serve to facilitate the solution of problems in maxima and 
minima: 

(1) If c is positive, f(x) and c X f(x) are maxima or minima 
for the same value of x; hence a constant positive factor or 
divisor may be rejected in finding this value of x. 

(2) logf(x) and/(z) are maxima or minima for the same 
value of x; hence log may be rejected. 

(3) c -\-f(x) and f{x) are maxima or minima for the same 
value of x; hence the constant c may be rejected. 

(4) If n is a positive whole number, [/(#)]" and f(x) are 

p 
maxima or minima for the same value of x; hence in [f(x)] q 

the denominator q may be rejected, or in Vf(x) the radical may 
be removed. 



MAXIMA AND MINIMA. Ill 



EXAMPLES. 



1. Find what values of x will render x 3 — 3a; 2 — 24a; -}- 85 a 
maximum or a minimum. 

Here f(x) = x 3 - 3x* - 24a; + 85, f'{x) = 3a; 2 - 6a; - 24, 
f"(x) — 6x — 6. Making /'(a;) = 0, we have 3z 2 — 6x — 24 = 0, 
the roots of which are x = + 4, x — — 2. Xow to determine 
whether these values of a; give maxima or minima values of 
fix), we substitute them for x in /"(a-). 

Thus: /""(4) = 6x4-6 = + 18, 

/"(- g) = 6 X - 2 - 6 = — 18. 

Hence, Art. 145, when x = 4, /(a:) is a minimum, and when 
x == — 2, /(a;) is a maximum. 

Let the student construct the locus oiy = x 3 — 3a; 2 — 24a; -f- 85, 
and thus exhibit these results graphically. 

2. Examine f(x) = x 3 — 3a; 2 + 3a; -f- 7 for maxima and 
minima. 

Here /'(.r) = 3z 2 - 6a; + 3, f"(x) = 6x - 6,f'"(x) = 6. The 
roots of 3a; 2 — Qx -\- 3 = are x = 1, a; = 1. Substituting these 
values of x mf"(x) and /'"(a;), we have/"(l) = 0,/'"(l) = 6. 
Therefore the function/(a;) has neither a maximum nor minimum 
value, which we also infer from the fact that the tivo roots of 
f(x) = are equal, Art, 140, Cor. I. 

3. Find the maxima and minima of x b — 5a; 4 -f- 5a; 3 — 1. 
Here f'{x) = 5a; 4 - 20a; 3 -f 15a; 2 ; .:f'(x) = is bx* - 20a; 3 

-f- 15a; 2 = 0, or (a; 2 — 4a; -f- 3)a; 2 = 0; the four roots of which are 
0, 0, 1, and 3. Rejecting the two equal roots, Art, 140, we have 
a x = 1, a 2 = 3, and since /'(— oo ) = 5(— qo ) 2 is -f, the given 
function is a maximum when x = 1, and a minimum when 
a; = 3. 

Therefore /(l) = is a maximum, 

and /(3) = — 28 is a minimum. 



112 DIFFERENTIAL AND INTEGRAL CALCULUS 

4. Examine (x — l) 4 (a;-|- 2) 3 for maxima and minima. 
Differentiating and reducing, we have 

The Foots of f'\x) --= are those of (x - l) 3 = 0, (x + 2) n = 
and Ix -j- 5 — 0; hence there are three roots each equal to 1, 
two each equal to — 2, and one equal to — f. 

Rejecting the two equal roots, we have a i = — f and a 2 — 1, 
and since /'( — co) = (— oo) 3 (— co) 2 ( — 7oo) is -f, f(x) is a max 
imum when x = — f , and a minimum when a = 1. 

5. Determine when J + c (x — a)* is a maximum or mini- 
mum. 

By (3) of Art. 146 we may remove b, by (1) c, and by (4) 3; 
hence we have /(a;) == (x — «) 2 ; ■'•/'(«) = 2 (a;— a), and a;— a = 0, 
or rtj = ^; and since /'( — qo ) is — J the given function is a min- 
imum when x = a. 

6. Find the maximum and minimum values of f{x) = y- , ( 8 « 

Here H*) = f±f • 

I. /'(a?) = gives x(x -\- 3) 2 = 0, of which one root is and 
the other two are — 3 and — 3. 

II. f'(x) = co gives (x + 2) 3 = 0, the three roots of which is 
each — 2. 

Rejecting the tivo equal roots, we have «, = — 2, a„ = 0; and 
since /'(— oo) is +, /'(— 2) = co is a maximum value of f(x), 
and f(0) = $£- is a minimum. 

7. If the derivative of f(x) is f(x) = x* - 10a; + 21, what 
values of x will render f(x) a maximum or minimum. 

The roots of x* — 10a; + 21 = are 3 and 7; substituting 
these for x in f"(x) = 2a; - 10, we have/"(3) = - 10 = - 4, 
and/"(7) = 14 — 10 = -J- 4; therefore /(3) is a maximum and 
/(7) is a minimum. 

Find the values of x which will give maximum and mini- 
mum values of the following functions: 



MAXIMA AND MINIMA. 113 

8. u — x 2 — Sx + 12. x = 4 3 min. 

9. u = x 3 — 3x 2 — 24x + 85. x = — 2, max. a; = 4, min. 

10. u = x 3 — 3a; 2 + 6.^ + 7 Neither a max nor min. 

11. u = 2x 3 — 21a; 2 4- 36x — 20. x = 1, max.. a: = 6, min. 

12. w = (a; — 9) 3 (.t — 8)\ a; = 8, max., a; = 8$, min. 

2^ — 7# 4- 6 

13. « = * — — x = 4. max.* a; = 16, min. 

a; — 10 . 

(x 4- 2) 3 

14. u — J ~ 2 . x = 3, max , x = 13, min. 

[x — 6) 

x 2 4- 3 

15. m = --. a; = — 1, max. ' x = 3. min. 

a; — 1 ' 

1 — x -j- x' . 

16. w = - — : ^ # = +, mm. 

1 -\- x — x 

ir . (a — x) 3 a 

iv 2* = -^-. 2;=-, mm. 

a — 2a; 4 

9 4 

18. u.= -~\- . a; = 9, max.; a; = 1|, min. 

a* o — a; 

19 w = sin a; { cos x. x = —, max. 

4 

7t 

20. w = sin a; (1 + cos a;), x = - , max. 

sin a; 7T 

21. u = - — —r -. a; = — , max. 

1 4- tan x 4 

x 

22. w = - — : 7 . x = cos a;, max. 

1 4- x tan x 

23. u = (l + a; 1 ) (7 — x) 2 . x—\, max.; x=0 and x=l, min. 

24. If x + y — 7i, what is the greatest possible value of xy ? 

in'. 
Make u = xy = x(n — x) = nx — x 2 . 

25. If y = mx -f- c, find the least possible value of Vx 2 -\- y 2 

c 
' Vrri 2 + 1 ' 
Make u = Vx 2 + # a = ^(1 4- w 2 )z 2 4- 2racrc 4- c 2 . 



114 DIFFERENTIAL AND INTEGRAL CALCULUS. 



26. A merchant bought a bolt of linen, paying as many 
cents for each yard as there were yards in the bolt, and sold it 
at 20 cts. per yard; required the greatest possible profit. $1.00. 

27. A club of x members has x 3 — 12ar -f 45a; -j- 15 dollars 
in its treasury; how much is that apiece if the amount is (1) 



a minimum ? (2) A maximum ? 



(1) $13.00; (2) $23.00. 



28. Find the value of cp when sin <fi — cos is a maximum. 

0= cos" 1 —'i^2 = 135°. 

29. Find the fraction that exceeds its square by the greatest 
possible quantity. -J. 

30. Find the fraction that exceeds its nth power by the 
greatest possible quantity. 



£)- 4 



31. Find a number x such that its rzth root shall be a max- 
imum. x = e = 2.7182 +. 

32. Find the altitude of the maximum rectangle inscribed in 
a given triangle. 

Let ABC be the triangle and HKFE the required maxi- 
mum rectangle; \etb=A B, h = CD, 
x=DG, and u — area of HKFE) then 

u = EF X DG = (EF)x. 

To express EF in terms of x, we 
have 

CD- CG .: AB : EF, 

or h- h -x::b : EF; 




.'. EF — y(h — x), and u — j-{h% — z~)t whose maximum value 
is required. 

Dropping j, we have f(x) = lix — x 1 , \ f'(x) — li — 2x = 0, 

whence x = --; that is, the altitude of the maximum rectangle 



is one half of the altitude of the triangle. 



MAXIMA AND MIXIMA. 



115 





33. Find the altitude of the maximum rectangle inscribed 
in a given parabola. 

Let BAG be the parabola, and IH 
the rectangle; let h = AD, x = AE, 
y — EH, and u = area of IH. 

Then y 2 = ±az, 

and u = 2(ED)(EH) = 2(h - x)y 
= 2{li -x)VIax i 

= ±V~ a (hx k -x*y 

Hence f{x) = hx* - x l , and f'(x) = ihx~* - %x l = 0, 
h 



I D K 

Fig. 21. 



or 



\* 



%Vx; whence x = ih; \ DE = jA. 



34. Find the altitude of a maximum cylinder with respect to 
its volume that can be inscribed in a given right cone. 

Let ED be the altitude of the cylinder 
inscribed in the cone DAG. Let BD = b, 
AD = h, ED = x, EF = y, and v = vol- 
ume of the cylinder; then 

v = 7t(EF) 3 ED = ntfx. 

To express y in terms of x, we have 

AD : BD : : AE : EF, or hi I : : h — x : y; 

whence 




y 



(h — x), and v 



nj^{h-x)x. 



.-. f(x) = (h - x)*x = h 2 x - 2hx z + x\ 

which is to be a maximum. 

f'{x) = ¥ - 4Jix + 3rc 2 = 0; whence Sx - li = 0, or x = p. 

That is, the altitude of the cylinder is -J of the altitude of 
the cone. 

35. Find the altitude of the cylinder in Fig. 22, if the cyl- 
inder is a maximum with respect to its lateral surface. 



116 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Denoting the lateral surface of the cylinder by S, we have 
S — 2n(EF)ED = 2rcyx = 27t-(h — x)x, which is to be a max- 
imum. Dropping the constant factor 2nj, we hava 

t (x) = 7ix — x 2 ; .• f''(x) = 7i — 2x = 0, or x = — 

36. Find the dimensions of a cylindrical open-top vessel 
which has the least surface with a given capacity. 

Let x = the radius of the base, y = the altitude,, s = the 
surface, and c — the capacity. 

Then c — 7ix 2 y ... (1) and s — nx 1 -j- 2nxij. ... (2) 

c 2c 
From (1), y = — - ; .*. s = 7tx* -j , which is to be a min- 

7tX X 



imum. 



,:/(x) = kx<+^, /'(*)= 2w - § = 0; 

X Jy 



V 4 and y = yi» 



whence x = y — , and «/ = y - , which is obtained by substitut- 

7t 7t 

insr for x in y = — =-. 

37. A rectangle is inscribed in a circle whose radius is R; 
find the sides of the rectangle when it is a maximum (1) with 
respect to its area, (2) with respect to its perimeter. 

Each side = R V2 in both cases. 

38. The hypothenuse of a right triangle is h; find the ratio 
of the other sides when the triangle is a maximum (1) with re- 
spect to its area, (2) with respect to its perimeter. 

Eatio = 1 in both cases. 

39. A cylinder is inscribed in a sphere whose radius is R; 
find the radius of the cylinder when it is a maximum (1) with 
respect to its volume, (2) with respect to its convex surface. 

(1) i VQR; (2) i V2R. 



MAXIMA AND MINIMA. 117 

40. A cone is inscribed in a sphere whose radius is E\ find 
the altitude of the cone when it is a maximum (1) with respect 
to its volume, (2) with respect to its convex surface. 

:i)|£; (2)4#. 

41. Find the maximum isosceles triangle with respect to its 
area that can be inscribed in a given circle. 

An equilateral triangle. 

42. Find the dimensions of a cone whic'i has the greatest 
volume with a given amount of surface. 

The slant height is three times the radius o± the base. 

43 Find the shortest distance from the point (x' = 1, y r = 

2) to the line 3y = ±x + 12. Ans. 2. 

44. Find the shortest distance from the poin-t {%' — 2, y' = 
1) to the parabola if — 4a?. tf% 

45. A square sheet of tin has a square cut out at each 
corner, find the side of the square cut out when the remainder 
of the sheet will form an open-top box of maximum capacity. 

A side = \ the side ol the sheet of tin. 

46. A man is at one corner of a square field whose sides are 
each 780 yards and wishes to go to the opposite corner in the 
least possible time; (1) how far along the side must he go before 
turning across the field if he can travel along the side and 
through the field at the rates, respectively, of 65 and 25 yards 
per minute ? (2) In what time will he reach the opposite 
corner? (1) 455 yards; (2) 40 min. 48 sec. 

47. Find the altitude of the least isosceles triangle circum- 
scribed about an ellipse whose semi-axes are a and b, the base of 
the triangle being parallel to the major axis. 36. 

48. A steamer whose speed is 8 knots per hour and course 
due north sights another steamer directly ahead, whose speed is 
10 knots and whose course is due west. What must be the 
course of the first steamer to cross the track of the second at 
the least possible distance from her ? N. (cos -1 f) W. 

49 If the statue of Washington on the cupola of the Capitol 
is a feet in height anH b feet above the level of an observer's 
eyes, at what horizontal distance from the centre of the cupola 



118 DIFFERENTIAL AND INTEGRAL CALCULUS. 
should the observer stand to obtain the most favorable view of 



the statue? \/b(a -j- b) feet. 

MAXIMA AND MINIMA OF FUNCTIONS OF TWO INDEPENDENT 

VARIABLES. 

147 Definition. — A function, u — f{x, y), of two independ- 
ent variables has a maximum or minimum value according as 

f(x + h,y + k) < f{x, y), or f{x + h, y + h) > f{x, y), 

for all small values of h and Jc, positive or negative. 

148. Conditions for maxima and minima. — In the function 
u =f(x, y) if we suppose x and y to vary simultaneously, it is 
obvious from Art. 139, that the maximum or minimum values 
of u will occur at the points where the total differential of u, 
[du], is equal to zero. That is, when 

[*] = S*+^ = d> 

As dx (= h) and dy (= h) are independent of each other, 
each term of (1) must be equal to zero. Hence 

j- = 0, and ¥ =0 (2) 

These equations express the first conditions essential to the 
existence of either a maximum or a minimum. 

Again, as u passes through a maximum or minimum value, 
[du] changes from + to — , or — to +, respectively ; therefore, 
in the former case [du] is decreasing, hence [d*u] is — , and in 
the latter [du] is increasing, hence [dht] is -f-. But the signs of 

^^ =J S dx ' + Sw dxd y+w df - - ■ (3) 

must evidently be independent of the signs of dx and dy, how- 
ever large or small these differentials may be supposed to be. 
This can be the case only when 

f^u\(£u\ ( d*u y 

\dx" l\dy* I > \dxdy) w 



MAXIMA AND MINIMA. 119 

For, making ^ = ^-, £ = ^. C = ^ , we have 



^- + 8£Ai + OT = ^ + ^ + (^-^ , 



(5) 



In order that (5) may preserve the same sign for all values of 
h and k, it is necessary that AC — B 2 should be positive ; for if 
negative, the numerator of (5) will be positive when k = 0, and 
negative when Ah -j- Bk — 0. Hence we have as an additional 
condition for a maximum or a minimum, AC > B 2 , or (4). 

With this condition, the sign of (5) depends on that of the 
denominator A. Hence for a maximum we must have 



d 2 u 
A or 

and for a minimum 



Ao *^<°> • ( 6 > 



Aor d^> < 7 ) 

It should be noticed that AC > B 2 requires that A and C 
should have the same sign. 

The exceptional cases, where B 2 = AC, Or where ^4 = 0, 
B = 0, (7=0, require further investigation, but we shall not 
consider them in this book. 

The conditions for a maximum or minimum value of 
u = f(x, y) are then, viz. : 

For either a maximum or a minimum, 

clu _ 3 du /rtX 

^ = 0, and ¥=0; . .... (8) 

, d 2 u d 2 u I d 2 u \ 

also d*df > \d^) (9) 

For a maximum, — < 0, j-~ < 0. ..... (10) 

For a minimum, ~^~ > 0, — > (11) 



120 DIFFERENTIAL AND INTEGRAL CALCULUS. 

EXAMPLES. 
1. Find the minimum value of u — x* -j- y 3 — 3axy. 

Here ^ = 3a; 2 - Zay; p- = 3y 2 - Sax; 

d\i d" 2 u d 2 u _ 

also ^=e.r, w = *y, a^/ = - 3a - 

Applying (8), we have 

x 2 — ay = 0, and y* — ax = ; 
whence x = 0, y = 0; and x = a, y = a. 

The values x = 0, y = 0, give 

cPu _ i% _ d 2 w 



dx* dy 2 dxdy 

which do not satisfy (9). Hence they do not give a maximum 
or a minimum. 

The values x = a, y = a, give 

d*u „ *? 8 w . d'u 

which satisfy both (9) and (11). 

Hence they give a minimum value of u, which is — a 3 , 

2. Find the minimum value of 

x* + xy -f # 3 — ax — by. $(ab — a 7 — b*). 

3. Find the maximum value of 

(a - x){a - y)(a + y - a). -^ 

4. Required the triangle of maximum area that can be inscribed 
in a given circle. The triangle is equilateral. 

5. Divide a into three parts, x, y, a — x — y, such that their 
continued product, xy(a — x — y), may be the greatest possible. 

x = y = a-x — y = -. 



MAXIMA AND MINIMA 121 

6. Divide 45 into three parts, x, y, 45 — x — y, such that 
#y (45 — x — yY may be a maximum. 

x = 10, ?/ = 15, 45 — x — ^ ■ = 20. 

7. Find the least possible surface of a rectangular parallelo- 
piped whose volume is a 3 . Qa'\ 

8. Find the dimensions of the greatest rectangular parallelo- 

x* if z 2 
piped that can be inscribed in the ellipsoid t + ?r + ^ = 1. 



CHAPTER VII. 

APPLICATIONS OF THE DIFFERENTIAL CALCULUS 
TO PLANE CURVES. 

TANGENTS, NORMALS, AND ASYMPTOTES. 

149. Equations of the Tangent and Normal. In Fig. 23 
let P(x lf y t ) be the point of tangency of the tangent TP; then 
the equation of TP is y — y \ = m(x — o^), where m is the 

tangent of the angle BTP. But, Art. 26, tan BTP = ^; 

therefore the equation of the tangent PT is 



y-^ = ^(*--*i)> ...... (A) 

cly cly 

where -^ is the value of ~~ with respect to the curve AP at 
ax 1 ax L 

the point (x x , y t ). 

Since the normal PN is perpendicular to the tangent or 

curve at P, its equation may be obtained from (A) by substitut- 

dx x , dy } , . , . 
mg — -=— s tor -p 1 , which gives 

y \ i 

y-y>=- j^( x - *J ( B ) 



EXAMPLES. 

1. Find the equations of the tangent and normal to the pa- 
rabola ?/ 2 == ±ax. 

122 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 123 

dy _ 2a m dy 1 _ 2a 
xxere ~^ — — » • • ~^ — • 

ax y dx x y l 

dy 
Substituting this value of -— in (A) and (B), we have 

V- V, = —(*—»,)* tangent; . . . . (1) 

y i 

y - y 1 = - |^(« - *,), normal. ... (2) 

2. Find the equations of the tangent and normal to the pa- 
rabola ?/ 2 = 18ic at the point x x = 2. 

Here 4# = 18 and y* = 18^; .*. 2a = 9 and #, = 6. 
Substituting in (1) and (2), and reducing, we have 

2y = 3x -f- 6, tangent, and 3y = — 2x -f- 22, normal. 

Find the equations of the tangent and normal to the follow- 
ing curves : 

3. The circle, %f + x* = R\ 

(1) yy x + xx, = B>; (2) yx t - xy x = 0. 

4. The ellipse, a 2 y 2 + &V = a*b\ 

(1) a*yy x + Vxx x = a'b*; (2) y-y x = ^(x - x x ). 

5. The cissoid, y'\2a — x) = x*. 

(1) tangent, y-y x = ± a ~ %l) ** (x - x t ). 

(2a - xj* 

6. Find the equation of the normal to y 1 — 6x — 5 at y x — 5. 

y = - ¥ + ¥• 



124 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



7. Find the equation of the tangent at y x = 4 to the cycloid 
x = 10 vers- 1 ^ - V20y-y\ 

y = 2x+ 20(1 -vers ^f). 

150. Length of Tangent, Normal, Subtangent, and Sub- 
normal. Let PT represent the tangent at the point P(x x , y x )> 
PA" the normal; then y l = FB, BT is the 
subtangent, and BN\s the subnormal. 
Let = the angle BTP, then 




T A B 

Fig. 23. 

that is, 

(2) 
that is, 

(3) 

that is, 

(4) 

that is, 



n (i) BT= BP cot = y 1 



dx l 



subtangent = y t 



dx x 
dy' x 



(0) 



BN = BP tan BPN = BP tan 0; 



subnormal = y J 



dx. 



TP - {BP + sin 0) = BP 



ds x 

¥7 ; 



tangent = ?/, 



<is. 



• • (D) 
(Art. 49) 

■ • (E) 



PJV r =(£P^cos0) = BP 



ds x 
dx~ ; 



normal = y x 



ds t 
dx % 



(F) 



In formulas (E) and (F), ds x = Vdx* -+- dy* 9 Art. 33. 

If the subtangent be estimated from the point T, and the 
subnormal from B, each will be positive or negative according 
as it extends to the right or left. 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 125 

EXAMPLES. 

1. Find the values of the subtangent and subnormal of the 
ellipse a 2 y 2 + 6V = c?b\ 



civ b^x 

Here —^ = ^-; substituting in (C) and (D) we have 

ax x a y x 



ci*v 2 x a ci* 

Subt. = jp- = — ; subn. = — 



¥x x x x a * 

2. Find the values of the subtangent and subnormal of the 
ellipse 9f + 4z 2 = 36, at x x = 1. 

Here a — 3, b = 2, x x = 1, which substituted in the preced- 
ing answers give subt. = — 8, subn. = — f . 

Find the values of the subtangents and subnormals of the 
following: 

3. y 3 = ax. Subt. = 3#, ; subn. = ^r— . 

4. Parabola, y* = ±ax. Subt. = 2x x ; subn. = 2a. 

5. y = a x . Subt. = ; ; subn. = a 2x i log a. 

3 log a & 



6. Find the length of the tangent of the tractrix, 



* = a log ( a+ ¥ * f ) - (a' - tff. Tang. = a. 

7. Find the lengths of the normal and subnormal of the 
cycloid x — r vers -1 — — V2ry — y 1 . 

Norm. = tf(2ry) ; subn. = \/(2ry — ?/ 2 ). 



126 DIFFERENTIAL AND INTEGRAL CALCULUS. 



151. Lengths of Tangent, Normal, Subtangent, and Sub- 
normal in Polar Co-ordinates. Let 
AP ( = s) be a curve, the pole, OP 
(= r) the radius vector, PT ti tangent 
at P. Let 6 = XOP. 

Draw OT perpendicular to OP 
and prolong it to meet the normal 
-* NP at N; then PT is the polar tan- 
gent, PN the polar normal, OT the 
polar subtangent, and ON the polar 
subnormal. Evidently, 




Fig. 24. 



(1) 
that is, 

(2) 

that is, 

(3) 

that is, 

(4) 
that is, 



ONP = OPT^ip. 
rclt 



dr 
r\W 



% 



(Art. 97) 
(Art. 98) 

v • (G) 



OT= OP tan OPT =r 
the polar subtangent - 
ON = OP cot ONP = r(^j) ; 
the polar subnormal = -jrr (H) 



TP= (OP -cos OPT) 



the polar tangent = -=— . 



dr 
ds~' 



(I) 



PN= (OP + sin ONP) = r 
ds 



rdd 
ds ' 



the polar normal = 



dir 



(J) 



In formulas (I) and (J) ds = Vdr 1 + rW, Art. 97. 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 127 

EXAMPLES. 

Find the tangent, normal, subtangent, and subnormal of the 
following polar curves : 

1. The spiral of Archimedes, r = ad. 

d£ = l m 
dr ~ a y 



r 2 
From (Gr), subt. = — ; from (H), subn. = a; 



from (I), tang. = - \/a 2 -\- r 2 ; 



from (J), norm. = Va? + r 2 . 
<l The logarithmic spiral, r = a 9 , 
dv 

Substituting in (G), (H), (I), (J), we And 



V T 

subt. = :j = mr; subn. = — ; 

log a m 



tang. = r V 1 -f m 2 ; norm. = r Vl + log 2 a. 

Find the subtangent and subnormal of the following: 

3. The hyperbolic spiral, rd — a. 

r 2 

Subt. = — a ; subn. = . 

a 

4. The Lemniscate of Bernouilli, r 2 = a* cos 26. 

^» /» a 

Subt. = — t—. — — -.; subn. = sin 20, 

a 2 sm 20' r 



128 DIFFERENTIAL AND INTEGRAL CALCULUS. 

152. An Asymptote to a curve is a tangent which passes 
within a finite distance of the origin and touches the curve at an 
infinite distance. A curve which has no infinite branch can 
have no real asymptote. 

In Fig. 23, let x Q and y Q represent the intercepts OT and OB. 
respectively; then, in (A), Art. 149, by making (1) y = and 
(2) x = 0, we find 

(i)'x. = z x -%^=OTi . . . . (K) 



(2)i /o = yi -x^ = OD . (L) 



Now, if the curve AP is of such a character that x or y , or 
both, remain finite when x l or y x , or both, become infinite (see 
Art. 154), the tangent TP will be an asymptote to the curve. 



EXAMPLES. 
1. Examine ?/ 3 = 6# 2 -|- x* for asymptotes. 

(Ill A.f I /y>^ fl II A-X 1 CJC 

Since -^ = — ^ — , -^ = — ] a ' , which substituted in 
dx y 2 dx 1 y; 

(K) and (L) give 

(1) x = x 1 - 



x x " r 



which = — 2 when x x — oo . 

Ax* + a?. 



(2) y, - y. fi . - t-jj + Ai» 



which = 2 when x x — oo . 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 129 

Therefore the straight line whose x and y intercepts are — 2 
and -1- 2, respectively, is an asymptote to the curve. 

Since the asymptote passes through the points (— 2, 0) and 
(0, 2), its equation is y = x -\- 2. 



153. General Equation of the Asymptote. Since the 
asymptote passes through the points (x , 0) and (0, y ), its equa- 
tion is 

y = dt {x ~ Xo) ' ' • " (M) ' or y = at x + *•■••■ W 

This equation enables us to determine whether or not any 
given curve has an asymptote, and, if it has, to find its equation. 

Let us denote the values which -^-- and y„ assume when x t = x 

by m 1 and &, , respectively; then we have 

y = m x x -f- ~b l (P) 

154. When the terms of the equation f(x, y) = are of 
different degrees, to find the relation of y to x when they are 
infinite, we may omit all the terms except the group which are 
of the highest degree with respect to x and y. 

Thus, when x is infinite, the equation ay* — ~bx 2 -\-cy-\-dx = e 

gives aif — Ix 1 = 0, or y = ± y —x. 

ct 

A curve like a 2 y 2 -j- b 2 x* = a 2 b 2 , or y* = x*(a' — x 2 ), etc., 
which has no infinite branch or branches, has no real asymptote; 
this is indicated by the fact that when x is infinite, y, as deter- 
mined above, will be imaginary. 

2, Find the asymptote of the hyperbola a'y 2 — Vx* = — a~b 2 . 

When x = x , y = ± -x, or y, = ± -x.. 
3 a Jl a l 



130 DIFFERENTIAL AND INTEGRAL CALCULUS. 



dx^a'yj y °-V> a* yi - y,' 

m 1 = ± - and ^ = 0, which substituted in (P) gives 



y — ± -x, Ans. 



3. Find the asymptote of the parabola y* = iax. 

When x = co , y = ± 2 Va#, or y l = ± 2 Vax y . 

dy 1 2a 2ax. . — _-.■ , _ 

jz- = ~ > Vo=y 1 - —r 1 = VfMx , which = co when x,= oo . 

Therefore the parabola has no asymptote. 

4. Find the asymptote of y 3 = ax 3 -\- x 3 . 

When x = go , we have y 3 = x 3 ; .'. y = x or y x = x t . 

dy y 2ax, 4- 3a:, 2 .. ax* 

~d^ = 3y ,' ' and y » = 3y; ; 

hence m 1 = 1 and 5, = Q , and the asymptote is y — x -}- ^- . 



155. Asymptotes Determined by Inspection. When an 
asymptote is perpendicular to the axis of x or y, it can often be 

dx 

determined by inspection. In the first case m l = oo , or -=-' — 0, 

which, substituted in (M), gives x — x 1 = 0, since, in this case, 
x — x 1 ; that is, if y , is infinite when x x is finite, x — x l = is 
the equation of the asymptote. 

x % 

Thus, in the cissoid, y 2 = , 

' J 2a — x 

y = oo when # — 2a; hence the line a; — 2a — 0, which is paral- 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 131 

lei to the axis of y and at a distance 2a from it, is an asymptote 
to the curve. 

Again, in xy = a or y = — , when x = 0, y = co ; therefore 

x 

x — 0, or the axis of ?/, is an asymptote to the curve. 

Similarly, in y — a x , when x — '— oo , y = 0; hence # = 0, 
or the axis of x, is an asymptote to the logarithmic curve. 

5. Find the asymptotes of xy — ay — bx — 0. 

(1) x - a = 0; (2) y - b = 0. 

6. Find the asymptote of y 3 = ax 2 — x\ 



i a 



a 
7. Find the asymptotes of y — c -f- 



y — c and x = b. 

8. Find the asymptotes of y 2 (x* + I) = x 2 (x 2 — 1). 

y=±x. 

9. Find the asymptotes of y*(x — a) = x 3 -\- ax 1 . 

x — a and y = ± (x -\- a). 

CURVATURE. 

156. A point moving along an arc of a curve changes its 
direction continuously, and the total change of direction is called 
the Total Curvature of the arc. 




The angle TtP', Fig. 25, through which the tangent PT 
rotates as the point of tangency P moves from P to P', being 



132 DIFFERENTIAL AND INTEGRAL CALCULUS. 

the total change of direction of the point P, is the total curva- 
ture of the arc PP'. 

157. Uniform Curvature. The curvature is uniform when, 
as the point of tangency moves over equal arcs, the tangent 
turns through equal angles; that is, when the distance described 
by the point varies as its direction. 

Let APm be the curve, AP = s, PP' = As, XEP = 0, 
Art. 49; then TtP' = Acp. Let PC and P'C be normals meet- 
ing at C. 

Supposing As oc Acp, we have (Art. 12) 

A \ Acp 

As — 7)i Ad), or — = —j- m 
7)1 As 

(1) Let us consider the meaning of —r- . If the distance As 

gives a total curvature of Acp, since As oc A<p, a distance of 1 

Acp . A(p 

will give a curvature of — r— . That is, — t— is the curvature 

per distance of unity, or the rate of change of the direction of a 
curve with respect to that of its length, for which reason it is 
called the curvature of the curve. 

(2) Let us determine the value of m. The circle is the only 
curve of uniform curvature. Hence, supposing As oc Acp, PP' 
is the arc of a circle whose radius (say R) is CP. The angle 
POP' = TtP' = Acp; but arc PP' = CP X angle POP'; that 
is, As = BAcp; hence m = R, and we have 

Acp 1 



As R' 

Cor. I. The curvature of any circle is equal to the reciprocal 
of its radius; and the curvatures of any two circles are inversely 
proportional to their radii. 

Cor. II. If R = 1, -j—= 1: that is, the unit of curvature 

As 

is the curvature of a circle whose radius is unity. 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 133 

158. Variable Curvature. When the curvature is variable, 
we define the curvature at any point P of the curve as the value 

which —j- would have were the curvature there to become uni- 

As 

form. Hence the curvature at P is the value of -=- at that point. 

' CIS 

159. Radius of Curvature. A circle tangent to a curve at 
any point, and having the same curvature as that of the curve 
at that point, is called the circle of curvature; its radius, the 
radius of curvature; and its centre, the centre of curvature. 

The curvature of this circle being that of the given curve, is 

equal to -=-; therefore the radius of curvature of APm at P, 
Fig. 25, is 

R = % 

d<p 



Cor. I. To express E in terms of the differentials of x and y. 

tan * = %> •■•t > = t ™- 1 %' 

hence we have 



d(p = <&+% ' also ' ds = {dx ' + df)i - 



. p *_ ds _{da? + dtff ( 1+ Ui) m 

• u - d<p~ ~~fa¥j ' or § • • (1} 

dx* 



* R will be positive or negative according- as the curve is convex or 
concave (Art. 173), but its sign is often neglected. 



134 DIFFERENTIAL AND INTEGRAL CALCULUS. 

EXAMPLES. 

1. Find the radius of curvature of the parabola y 2 = 4a#. 

Hers %L - 2? and *-£---¥£. 

Here & ~ y' an(l rfaj- ~ ^ 

Substituting in (1), we have 

At the vertex, where y = 0, we have R = 2a, which is evi- 
dently the minimum radius of curvature. 

2. Find the radius of curvature of the ellipse a*y* -f- b*x* = 
a*b\ 

dy___Vx <py _ b> 

dx ~ a?y y dx x ~ tfy* ' 

b* [ aY J ~ a A V 



At the vertex x = a, y — 0, R = — , and at the vertex x = 0, 

a 2 
y = b, R = j-> which are respectively the minimum and maxi- 
mum radii of curvature. 

3. Find the radius of curvature of the cycloid 



■xV 



x = r vers -1 — — v%ry — y' 



Here - = —=M= ■ ■ 1 + ^! = ^. 

dy V2ry - y> ' dx' y » 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 135 
d*y r 



dx 2 



. \ R = 2 f 2ry, 



which equals twice the normal. 

4. Find the radius of curvature of the logarithmic curve 

y = *". ^ = (m 2 + y 'f 

my 

5. Find the point on the parabola y* = 8# at which the 
radius of curvature is 7|§. y = 3, x = 1-J. 

6. Find the radius of curvature of y = x* — 4x 3 — l&c' at 
the origin. „ _ 1 

K ~ 36 * 

7. Find the curvature of the equilateral hyperbola xy = 12 
at the point where x = 3. 1 __ 24 

R ~125* 

8. Find the radius of curvature of the catenary 



/ x - -\ 



* = £ 



9. Find the radius of curvature of the hypocycloid 

x l + y 1 = a* A' = 3(<m/)*. 



160. The radius of curvature in polar co-ordinates can 

be found by transforming the value of R given in the answer to 
Ex. 7, Art. 112, to polar co-ordinates. We thus obtain 

b =^-^-wf (*) 

r 4- 2 r 



136 DIFFERENTIAL AND INTEGRAL CALCULUS. 



EXAMPLES. 

1. Find the radius of curvature of the spiral of Archimedes, 
r = ad. 



-rr dr a?r 

Here M =a > W = 0i 



substituting in (2), we have 

(r 2 + a 2 ) 1 = a(l + fr) 1 
r 2 + 2a 3 2 + 2 * 

Find the radius of curvature of the following: 



2. The logarithmic spiral r = a 9 . R — r Vl -\- (log a)' 



9 



3. The cardioid r = «(1 — cos 0). P = - \ y 2ar. 

o 



a a 



4. The lemniscate r 2 = a 2 cos 20. R = —~ , 

3r 



CONTACT OF DIFFERENT ORDERS. 

161. Let ?/ =f(x) and ?/ — 0(a) be any two curves referred 
to the same axes. Let the curves intersect at the point P, 
whose abscissa is a, then f(a) = 0(a). If f{a) = 0(a), and 
/'(a) = 0'(a), the curves are tangent at P, and are said to have 
a contact of the first order. If f(a) = 0(a), /'(a) = 0'(a), and 
f"(a) = 0"(a), the curves have the same curvature at P, and 
their contact is of the second order. If, in addition, /'"(a) = 
0"'(a), their contact is of the third order; and so on. Thus, 
contact of the nth. order imposes n -f- 1 conditions. 

162. Two curves cross or do not cross at their point of con- 
tact, according as their order of contact is even or odd. 

Let x = a be the abscissa of the point of contact of the 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 137 

curves y = f{x) and y = <p(x), then f(a) = <p{a). Let h be a 
small increment of x. By Taylor's formula, we have 



f(a + A) =/(«) +/»A +f"(a)~ +f'"(a)^+; (1) 



0(a + A) = <p(a) + 0'(a)h + 0"(«)y + 0"'(«)-jj +• (2) 



Subtracting (2) from (1), we obtain 
f(a + h)-<p(a + A) = A[/»-0»] + ~[f"(a) - <fi"(a)] 



+ !"[/'"(«) - <£"»] + ^"[/' v («) - *"(«)] +• (3) 



(«) If /(») — 0(ic) changes sign as x increases from a — hto 
a-\-h, the two curves evidently cross at a; if not, the curves 
touch each other, but do not cross. 

(b) If the contact is of an odd order, the first term of the 
second member of (3), which does not vanish, contains an even 
power of h; hence the sign of the second member, and therefore 
the first, undergoes no change as x passes from a — h to a -f- li, 
and the curves do not cross. 

(c) If the contact is of an even order, the first term of the 
second member of (3), which does not vanish, contains an odd 
power of h; hence, in this case, f\x) — <p(x) changes sign as x 
passes from a — h to a -f li, and therefore the curves cross. 

Cok. I. At a point of maximum or minimum curvature, the 
circle of curvature has contact of the third order with the curve, 
for it does not cut the curve at such a point. 

Cor. II. If two curves are tangent to, and cross each other at, 
a certain point, they have contact of at least the second order. 



138 DIFFERENTIAL AND INTEGRAL CALCULUS. 

EXAMPLES. 

1. Find the order of contact of the two curves 

y = x 3 — 3£ 2 -f 7 and y + 3z = 8. 

By combining the two equations we find that (x = 1, y = 5) 
is a point of contact. 

Making /(«) = x* — 3# 2 + 7 and <f)(x) — 8 — 2>x, we have 

/'(a) = 3s 1 - 6s, 0'(s) = - 3; .'. /'(I) = 0'(1) = - 3; 
/"(a?) = te - 6, 0"(s) =0; /. /"(I) - 0"(1) = 0; 

/'"{*) = 6, 0'"(s) = o; .'./'"(I) > 0"'(1). 

Hence the contact is of the second order. 

2. Find the order of contact of the parabola y* = 4z and the 
line 3y = x + 9. First order. 

3. Find the order of contact of the curves 

y = 3x — x 1 and xy = 3x — 1. Second order. 

4. Find the order of contact of 

y = log (x — 1) and z 2 — 6# + 2y + 8 = 0, 

at the point (2, 0). Second order. 

5. Find the order of contact of the parabola y 2 = 4x + 4 and 
the circle # 2 + # 3 = 2a; + 3. Third order. 



163. Osculating Curves. The curve of a given species 
that has the highest order of contact possible with a given curve 
at any point is called the osculating curve of that species. 

A curve may be made to fulfil as many independent condi- 
tions as there are arbitrary constants in its equation, and no 
more. Therefore, in order that y —f(x) may have contact of 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 139 

the nth. order with a given curve at a given point, the equation 
must involve n -J- 1 arbitrary constants. 

Hence, as y = ax -J- 5 has two constants, the osculating 
straight line has contact of the first order. 

As (x — a) 7 -f- (y — by = r 2 has three constants, the osculat- 
ing circle has, in general, contact olthe second order. 

164. To find the osculating straight line at any point 
{x' y y') of a given curve y = f{x). 

The equation of a line is 



y = ax + b (1) 

Since the line and curve pass through {x\ y'), we have 

y' = aaf + b=f&). ...... (2) 

Also, % = a =/'(*'), ...... (3) 

since/'(z') = 0' (ab- 
solving (2) and (3) for a and b, we have 

.=!£, and t = f-&*. 

which, substituted in (1), gives 

Therefore the osculating straight line is a tangent to the 
curve, as would be inferred. 

165. To find the radius of the osculating circle at any 
point of a given curve, y =f(x). 

The general equation of a circle whose radius is r is 

(x-ay + (y-by = r> (1) 



140 DIFFERENTIAL AND INTEGRAL CALCULUS. 
Differentiating twice successively, we have 

X - a+{ y-b)<^L=0, ...... (2) 



i + d £+^-^=° < 3 > 



From (3), 9 -i=-**+J£ (4 ) 

*»t». * — J^|P* (5) 

Substituting (4) and (5) in (1), we have 

By comparing this result with formula 1, Art. 159, it will he 
seen that the osculating circle is the same as the circle of cur- 
vature. 



INVOLUTES AND EVOLUTES. 

166. An Involute may be regarded as a curve traced by a 
point in a thread as it is unwound from another curve, called 
the Evolute. 

Thus, imagine a thread stretched around the curve A 1 P i m l 
with one end fastened at m l ; if the thread is unwound by carry- 
ing the point at A above and around to the right, that point of 
the thread will trace the involute APm of which A l P x m l is the 
evolute. 

An evolute may have an unlimited number of involutes, for 
A may be any point on the curve A x m v 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 141 

111 what follows the chief object is to deduce certain prop- 
erties of the evolute from its involute, or vice versa, and for 
uniformity the co-ordinates of P (the involute) will be repre- 
sented by x, y, and those of P x (the corresponding point of the 




evolute) by x x , y x ; the arc AP by s; the arc AP X by s r ; and the 
angles of direction of AP and AP X , at P and P x , by and cp x , 
respectively. 

167. Elementary Principles. I. PP X — the arc AP X = s x . 




II. PP X is tangent to AP x m x at P x , for it has the same 
direction as the curve at that point. 

III. The line PP, is a normal to the curve APm at P. 



142 DIFFERENTIAL AND INTEGRAL CALCULUS. 
For, draw TP tangent to the curve AP at P. . 

(P^ + tPP^^P) 2 , or ( yi -yy + {x - Xi y = Sl \ (i) 

(#i - y)(dyi - dy) + (x - £,)(dz - dxj = s.ds,. . (2) 
Again, P X E = P,P sin EPP t , or y x —y = s ip; . . (3) 

also, PP = P 2 P cos PPP 1 , or a? — as, = - s^ 1 . . (4) 

Substituting in (2) from (3) and (4), and reducing, remem- 

ciii doc 

bering that dx* -j- dy* = t^ 2 , we have -~ = — — ; that is, 

tan X = - cot ; .-. X = | + 0, or PFX = f + P^FX; 
hence P X P is perpendicular to the tangent PT, 

Cor. I. Since sin <fi x = cos 0, -^ = — ; . . . . (5) 

cts . cis 

also, since cos 0, = — sin 0, -j- 1 = — ~jj- . . . . (6) 

Cor. II. The point P, is the centre of curvature of the curve 
^LPra at P. 

For, if circles be described from P/ and P, as centres with 
P/P' and P, P as radii, respectively, the arc PP' will lie within 
the one circle and without the other, since the straight line 
P/P' is equal to the partly curved line P X P X P. Hence the 
circumference of the circle whose centre is P, crosses and touches 
the curve APm at P (Art. 162, Cor. II). 

Cor. III. Since P X P = s z = P, we have (Art. 159)' 

s = cU = (rZ* 2 + <fy 2 )* 
1 d0 ^fy ' ^ ' 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 143 
Coe. IV. From (4) and (6), x 1 = x- sf^, .... (8) 

(lor 
and from (3) and (5), y l = y -f- s 1 ^-. .... (9) 

CIS 

Substituting for s x in (8) and (9), from (7), we have 

lx + d J-\ d JL 1 + *£ 

.*• = * j* ' and y. = ^ + -^- -(io) 



ofo 2 cfa 2 

These values of x 1 and f/j are the values of the co-ordinates 
of the centre of curvature at the point P. 

168. To find the equation of the e volute of any given 
curve. 

By differentiating the equation of the given curve, and sub- 
stituting the results in (10), x l and y 1 may be expressed in terms 
of x and y. If, between the equations thus obtained and that 
of the given curve, x and y be eliminated, the resulting equation 
involving x 1 and y 1 will be the equation of the evolute. 



EXAMPLES. 
1. Find the equation of the evolute of the parabola y* = lax. 

Here %._?«. <%L = _W 

dx y ' dx* y 3 ' 

Substituting in (10), we have 

, y 2 + 4a 2 2a if ' ," 

x t = x+ 2— El ^_ = 3x + 2a; 

y 1 y 4a 2 

_ x' — 2a m 
x — - ; 



144 DIFFERENTIAL AND INTEGRAL CALCULUS. 



f + 4a 2 y> f 

f 



Vl y if 4a 2 4a a 



These values of x and y substituted in y 1 — ±ax give 

which is the equation required; hence the evolute is the semi- 
cubical parabola. 

Cor. I. The length of the arc of the evolute s 1 may be found 
by formula (7), Art. 167. 

2. Find the evolute of the ellipse ay + 5V = a 2 b\ 

{ax^f + {h yi f = {a' - ^) § . 

3. Find the co-ordinates of the centre of curvature of the 
cubical parabola y 3 = tfx. 

_ a" + 15 y* _ a*y-$ij % 
Xl ~ 6a*y > Vx ~ ' 2a* ' ' 

4. Find the co-ordinates of the centre of curvature of the 

X _x 

catenary y = ~(e a + e a ). x ^ =x ~\ *V ~ ^ & = 2 ^ 

—J 5. Find the co-ordinates of the centre of curvature, and the 
equation of the evolute, of the hypocycloid x* -f y* = a?. 

Xl = x + 3 Vxjf, y, = y + dj/xY; fa + V x f + fa - V,f 
- 2a?. 

6. Find the evolute of the equilateral hyperbola xy = m 2 . 

fa + #i) § ~ fa - ^) § = ( 4 "0 § » 

Note. — First prove that 

m fm , xV .. m fm x V 

1 ' yi 2\x l 7n/' 1 * l 2 \x ml 

and thence derive the equation of the evolute. 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 145 

ENVELOPES. 

169. Let 

f(x,y,a)=0 (1) 

be the equation of a curve, a being some constant quantity. If 
we assign different values to a, we will obtain a series of distinct 
curves, but all belonging to the same system or family of curves. 
One of the curves of this family can be obtained by increasing a 
by h, thus converting (1) into 

/I*,y,« + A) = (2) 

If li be supposed indefinitely small, the curves (1) and (2) 
are said to be consecutive. 

The points of intersection of the curves (1) and (2) approach 
definite limiting positions as h approaches 0, and the locus of 
these limiting positions, as different values are assigned a, is 
called the Envelope of the system f(x, y, a) = 0. 

The quantity a which remains constant for any one curve of 
the series, but varies as we pass from one curve to another, is 
called the variable parameter of the series. 

170. The envelope of a series of curves is tangent to every 
curve of the series. 

Let A, B, C be any three curves of the series, A and B inter- 
secting at P, and B and C at P' . 




Fig. 28. 



As these curves approach coincidence, the limiting positions 
of P and P' will be two consecutive points of the envelope and 
of the curve B. Hence the envelope touches B. 

As an illustration see example 1 under the next article. 



146 DIFFERENTIAL AND INTEGRAL CALCULUS. 

171. To find the equation of the envelope of a given 
series of curves. 

The point of intersection of (1) and (2) will be found by 
combining the equations. Now, subtracting (1) from (2), we 
have 

fix, y, a + h) -f{x y y f a) _ A 

h ~ U {6} 

When the curves approach coincidence, h approaches 0, and 
(3) becomes 

^«)=0 W 

Thus, equations (1) and (4) determine the intersection of 
any two consecutive curves. Hence, by eliminating a between 
(1) and (4), we shall obtain the equation of the locus of these 
intersections, which is the equation of the envelope. 



EXAMPLES. 
1. Find the envelope of y = ax -| , a being the variable 

parameter. 

iix 
y — ax -f- — is the equation of a line, as MN, Fig. 29. 

tt 

When a receives an increment It, the line takes a new position, 
say M'N', which intersects the former line at c. As h 
approaches 0, c approaches p, a point on the locus (APm) of all 
similar intersections. 

Differentiating with respect to a, x and y being constants, 
we have 

= x -', whence a = ± \ — . 

y = ± [ Vmx -\- Vmx], or y* = ±mx. 
which is the equation of a parabola, 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 147 



Let it be observed that the problem is the same as that of 



m 



finding the curve of which y = ax -} is the tangent. 

0/ 




Fig. 29. 



2. Find the curve whose tangent is y — mx -|- a Vni 2 -J- 1, m 
being the variable parameter. x 2 -f- ?/ 2 = a 2 , a circle. 

3. If a right triangle varies in such a manner that its area is 
constantly equal to c, find the envelope of the hypothenuse, or 
the curve to which the hypothenuse is the tangent. 

Let OA — a, OB = b; then the equation of AB is 



b 



-K--1. 



(l) 



But ab = 2c, or b = 



2c 



X - _L W - 1 

a " h 2c"■ l, 



(2) 



where a is the variable parameter. 

/2cx 
Differentiating (2), we get a = y , which, substituted in 

(2), gives ay = ^-, an equilateral hyperbola. 

2 



148 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Solve the preceding problem on the hypothesis that the hy- 
pothenuse is constantly equal to c. 

^ + V* — c*> the hypocycloid. 
Since a normal to a curve is tangent to the evolute of the 
curve, the latter is the envelope of the successive normals, or the 
locus of their intersections. 

4. Find the evolute of the parabola y 2 = 4##, taking the 
equation to the normal in the form y = m{x — 2a) — am 9 , m 
being the variable parameter. 27ay 2 = 4(# — 2a)\ 

5. Find the evolute of the ellipse a 2 y 2 + # 2 ^ 2 = a 2 b% taking 
the equation of the normal in the form 

oy = ax tan a' — (a 2 — ¥) sin a', 

where the variable parameter a' is the eccentric angle. 

(ax)* + (byf = (a' - b 2 f. See Ex. 2, Art. 168, 

6. Find the curve whose tangent is y = mx -f- (a' 2 m 2 -f b 2 )$, 
m being the variable parameter. a 2 y 2 -f- Vx 2 = a 2 b 2 . 

7. Find the envelope of the family of parabolas whose equa- 
tion is y 2 = a(x — a). y = ± \x. 

8. Find the locus of the intersections of x cos a -f- y sin a = p 
with itself as a increases continuously. x 2 -f- y 2 = p 2 - 

9. Find the envelope of all ellipses having a common area 
(7rc 2 ), the axes being coincident. xy — ± \c 2 . 

10. Find the evolute of the curve x* -\- y* = a*, the equation 
of whose normal is y cos a' — x sin a' = a cos 2a f , where a f is the 
angle which the normal makes with the axis of x. 

(x + y f _j_ ( x - y f - 2a\ See Ex. 5, Art. 168. 

11. Find the equation of the curve, the equation of its tan- 
gent being y = 2mx -f- in\ where m is the variable parameter. 

TRACING CURVES. 

172. The Rudimentary Method of tracing a curve is to 
reduce its equation to the form of y = f(x) ; that is, solve the 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 149 

equation f(x, y) = for y, assign values to x, find the correspond- 
ing values of y, draw a curve through the points thus determined, 
and it will be approximately the curve required. This process 
is laborious, and often impossible on account of our inability to 
solve f(x, y) = for y. 

The General Form of a curve is usually all that is desired, 
and this can generally be found by determining its singular or 
characteristic points and properties, and these are embraced 
chiefly in the position of certain" turning-points of the curve, 
the direction of curvature between these points, and. where and 
how the branches intersect or meet each other. In addition, we 
may find, by previous methods, where the curve cuts the axes, 
whether or not it has infinite branches, asymptotes, etc. 

173. Direction of Curvature. — The terms Convex and 
Concave have their ordinary meaning when applied to the arcs 

of curves. 

Thus, as seen from some point 
below, the arcs AB X and OB are 
concave, and B x and BE convex. 
174. A Point of Inflection 
is the point at which the curve 
changes from concave to convex, 
or from convex to concave ; as the 
points B J} C, B. 

Principles. — The slope (-r-) of the curve evidently de- 
creases as the point P(x, y) moves from A along the curve to 
the right until P reaches B t , and then increases until P reaches 

(7, etc. Therefore (1) when the arc is concave, ~— decreases as 

x increases, hence (Art. 25) its derivative -y~r is — ; (2) when 

,, dy . . , d*y . . 

the arc is convex, -f- increases as x increases, hence -3-5 is 4-. 

' dx dx 2 

Therefore, I. At any point of the curve y =f(x), the curve 
is concave or convex according as -^~ is negative or positive. 




150 DIFFERENTIAL AND INTEGRAL CALCULUS. 

d?y dv 

II. The roots of -~ = or oo which will render —• a maxi- 
ma; 2 dx 

mum or minimum are the abscissas of the points of inflections. 



EXAMPLES. 

Examine the following curves for concave and convex arcs, 
and for points of inflection. 

1. y = x * - 6x + 7. p? = 2. 

Since ~^~ is +, the curve is convex at every point. 

2. y = re 3 - 6a; 2 + 17a; - 6. |^ = 6(a - 2). 

The root of x — * 2 = is 2, the point of inflection; the curve 
is concave when x < 2, convex when x > 2. 

3. y = x* - 12a; 3 + 48a; 2 - 50. 

Points of inflection, a; = 2, a; = 4; curve convex when x < 2 
and > 4, and concave when 2 < x < 4. 

— x ~ 2 ^y 2_ 

y " a; - 3' dx* " (a; - 3) 3 ' 



Point of inflection at a; = 3; convex when a; > 3, concave 
when x < 3. 

6.y = log(*-l). _ = _-__ 

= oo gives (a; — l) a = 0, which has two equal roots; 



hence, Art. 140, there is no point of inflection; curve concave 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 151 



6. Prove that the curve y = — — — - has points of inflection. 

ct ~\~ x 

at (0, 0), (a V3, la VI), {-aVd,-ia 1/3). 

7. Prove that the witch of Agnesi, x*y = 4a i (2a — y), has 
points of inflection at (± \a Vo, |«), and is concave between 
these points and convex outside of them. 

8. Find the points of inflection of y = sin 2x + cos 2x. 



SINGULAR POINTS. 

175. The Singular Points of a curve are the turns and 
multiple points. 

A Turn in rectangular co-ordinates is a point at which a 
curve ceases to go (1) up or down, or (2) to the right or left, 




Fig. 31. 



and begins to go in the opposite direction. The former, as 
B, E, F, G, are called y-turns, and the latter, as 0, D, H, 
x-turns. 

The x-turns and y-turns evidently occur at the maximum or 
minimum values of x and y, respectively. 



152 DIFFERENTIAL AND INTEGRAL CALCULUS. 

A Multiple Point* is one through which two or more 
branches of a curve pass, or at which they meet. A multiple 
point is double when there are only two branches; triple when 
only three, and so on. 

A Multiple Point of Intersection is a multiple point at 
which the branches intersect (Fig. 32, a). 




c e 

Fig. 32. 

An Osculating Point is a multiple point through which two 
branches pass, and at which they are tangent (Fig. 32 ? b, c). 

A Cusp is a multiple point at which two branches terminate 
and are tangent (Fig. 32, d, e). A -cusp or osculating point is 
of the first or second species according as the two branches are 
on opposite sides (Fig. 32, b, d) or the same side (Fig. 32, c, e) 
of their common tangent. 

A Conjugate Point is one that is entirely isolated from the 
curve, and yet one whose co-ordinates satisfy the equation of the 
curve. 

For example, in the equation y = (a + x) Vx, if x is negative 
y is imaginary, yet the co-ordinates of the point (x = — a, y = 
0) satisfy the equation. Hence (— a, 0) is a conjugate point. 
A conjugate point is, generally, the intersection or point of 
meeting of two imaginary branches of the curve, and may, in 
exceptional cases, also lie on a real branch of the curve. 

There are other singular points, such as Stop Points, at 
which a single branch of a curve stops suddenly, and Shooting 
Points, at which two or more branches stop without being tan- 
gent to each other. But as these rarely occur, they are omitted 
in this book. 

* See Taylor's Calculus. 



co ; . 


dy 


0; . 


ax 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 153 

176. To determine the positions of the singular points of 
a curve. 

Let u =f(x, y) = be the equation of the curve, free from 
radicals. Then (Art. 109) 

du 

dy _ dx 

dx du 

dy 

(a) For the z-turns, -f- 

v ' dx 

(b) For the ?/-turns, -p 

(c) For multiple points, ~i~, by definition, has two or more 

dy 
values ; hence, since u contains no radicals, ~ must be of the 

dx 

form -. Therefore 

du , du A 

t=- = and -r- = 0. 

/XT/, 

Hence, to find the a;- turns we have u — and ^— = ; to 

dy 

du 
find the y-turns, we have w = and — = 0; and the values of 

y and x which satisfy all these equations are the co-ordinates of 
the multiple points. 

177. To determine the character of the multiple points 
of a curve. 

From the definitions of the multiple points it follows that: 

I. At a multiple point of intersection -j- has two or more 

unequal real values. 

du 

II. At an osculating point or a cusp y- has two equal values. 

III. At a conjugate point at least two of the values of -j- are 
imaginary. 



154 DIFFERENTIAL AND INTEGRAL CALCULUS. 



EXAMPLES. 
Find the singular points of the following curves. 
1. u = x 2 - xy + if - 3 = 0. . 



du 

dy 



(1) 



x + 2y = 0; 



du 



(2) ^ = 2x-y = 0. . . (3) 



From (1) and (2) we find (2, 1), (— 2, — 1), the co-ordinates. 
of the #-turns A and A'. 

Y 




Fig. 33. 

From (1) and (3) we find (1, 2), (— 1, — 2), the co-ordinates 
of the y-turns B and B'. 

Since neither pair of these values satisfies (1), (2), (3), the 
curve has no multiple points. 
2. u = Aif - (25 - x 2 )(x 2 + 7) = 0. 




du 
dx 



= 2x{x i + 7) - 2z(25 - x 1 ) - 0. 



Fig. 34. 



From these equations we find 

(a) the z-turns, (5, 0), (— 5, 0), 

(±•=7,0); 

(b) the y-turns, (3, ± 8), (0, ± f V7), 

(- 3, ± 8). 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 155 



The figure (34) is only an approximate representation of the 
curve. 

z-turns, (± 4, 18), and (± A ^dyf, ^. 



y-turns, (0, 4), (0, 0), and 
:y 5 = 0. 
= a(2x' > - 3# 2 ) = 0. 



± g vw, 19^). 



w = a: 4 -f- 2ax 2 y 

du 
dy 

du 



dx 



ix(x 2 + ay) 



0. 



\ y-turns, (0, 0), (a, — a), (—a, —a); 




Now there appears to be an #-turn and a y-turn at the point 
(0, 0), and in a certain sense this is evidently true; but we should 
regard the result as signifying that (0, 0) is a multiple point of 
some hind, since x = 0, y = satisfy equations (1), (2), and (3). 

Let us now determine the character of the point. Dividing 
(3) by (2), we have 

dy _ 4# 3 + ^cixy 
dx Say 2 — 2 ax*' 

Our object now is to find the value of the slope -~- at the 

dti 
multiple point (0, 0). For these values of x and y, -j- assumes 

the form of -, hence the value required may be obtained by 
Art. 137. 



156 DIFFERENTIAL AND INTEGRAL CALCULUS. 



We see from Ex. 5, Art. 137, that -f- = and ± V2 at the 

ClX 

point (0, 0). Hence the origin (0, 0) is a triple point, the three 
branches which pass through the point being inclined to the 
a-axis at the angles 0, tan -1 V% and tan -1 (— 4^2), respectively, 
as in the figure. See Art. 179. 

5. y 2 = «V - x\ 

z-turns, (0, 0), (a, 0), (— a, 0); 

y-tnrns, (0, 0), g-Ya/ ±| «'),(- | V% ± i-« J ). 
The point (0, 0) is a double point of intersection, since at 



that point 



dy^ 
dx 



±a. 



6. Examine y 2 (ci 2 — x*) — x x = for multiple points. 

At the point (0, 0), ~~ = ± 0; that is, it has two equal 
ax 

values; hence (0, 0) is an osculating point or a cusp; and since 

the curve is symmetrical with respect to both axes the point is 

evidently an osculating point of the first species. 

7. Determine the general form of the curve y 2 = a 2 x\ 
When x = oo , y = ± oo ; hence the curve has two infinite 

branches, one in the first and one in the fourth quadrant. 

When x is negative, y is imaginary; 
hence the curve does not extend to the 
left of the #-axis. 

When x = 0, y = 0; hence both 
branches start from the origin. 

At the point (0, 0), -|-= ± 0; hence, 

since the curve is symmetrical with 
respect to the z-axis, the origin is a 
cusp of the first species. 




Fig. 36. 



DIFFERENTIAL CALCULUS AND PLANE CURVES, 157 



Again, since -^ = 



3a 



-., the upper branch is convex and 




Fig. 37. 



2Vx 
the lower concave. 

8. Examine the curve (y — ar 2 ) 2 = x h , or y = x* ± x*. 

Has two infinite branches, one in the first and one in the 
fourth quadrant, both starting from the 
origin. For every positive value of x, y 
has two real values, both of which are posi- 
tive as long as x < 1, but at the point 
where x = 1 the lower branch crosses the 
z-axis. The origin is a cusp of the second 
species. 

178. Tracing Polar Curves. Let 
f{r, 6) = be the polar equation of the curve. 

(a) By solving the equation f(a, 6) = for 6, we find the 
direction of the curve at the point r = a. If a = 0, the values 
of 6 will be the angles at which the curve cuts the polar axis at 
the pole. 

dv 
{b) By solving the equation -^ = for we find the values 

of for which r is a maximum or minimum, or the r-turns, at 
which the curve is perpendicular to the radius vector. 

9. Trace the curve r = a sin 3#, Fig. 38. 

{a) Making r — 0, we have sin 36 = 0; hence 6 = 0, \n, \n, 

which are the angles at which the curve cuts the polar axis at 

the pole. 

civ 
(b) -jT2 = 3« cos 36 = 0; hence the values 




d6 



TC 



of 6 at the r-turns are \n i — ,- |-7r, at which 



Fig. 38. 



points r ■■ 
Since 



dr 

d6 



-a, -{-a, respectively. 

3a cos 36, r increases from to 



a, while 6 increases from to \n\ r decreases from a to — a, 
while 6 increases from \n to \n\ r increases from — a to -f- a, 
while 6 increases from \n to f 7t; and r decreases from a to 0, 



158 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



while increases from fzr to n. Further revolution of the 
radius vector would retrace the loops already found. 

10. Trace the curve r = a sin 20,. Fig. 39. 
From this and the previous example, we 

infer that the locus of r = a sin nd consists of 
n loops when n is odd, and 2n loops when n 
is even. 

11. Trace the curve r = a cos cos 20, or 
Fig. 39. r = fl ( 2 cos 3 - cos 0), Fig. 40. 

12. Trace the lemniscate r 2 = a? cos 20, Fig. 41. 






'0 
Fig. 41. 

179. The character of multiple points in rectangular co- 
ordinates may often be more easily determined by changing to 
polar co-ordinates, and applying (a) of Art. 178. 

Thus, in Ex. 4, Art. 177, make y = r sin and x = r cos 0, 
divide by r 3 , and we have 

r cos 4 + 2a cos 2 sin - a sin 3 = 0. 

Now making r = 0, and we have 

sin 0=0, and tan 2 = 2; 

that is, the angles at which the curve cuts the #-axis at the 
origin are sin -1 0, tan -1 V2, tan -1 — V2. 
Trace the following curves: 

13. The Cissoid, y\2a - x) = s\ 

14. The Conchoid, xy = (6 2 - ?/ 2 )(rt + y)\ 



DIFFERENTIAL CALCULUS AND PLANE CURVES. 159 



15. 


The Witch, 


(x 2 + ±a 2 )y = 8a\ 


16. 


The Lituus, 


rV~6 = a. 


17. 


The Parabola, 


2 
r = a sec - . 

Z 


18. 


The Curve, 


• 3# 

r — flsm -. 



19. The Cardioid, r = a(l - cos 6). 

20. The Hypocycloid, x* -f y % = a § . 

21. Examine a?/ 2 = a; 3 — bx 2 for multiple points. 

(0, 0) is a conjugate point. 

22. Prove that y 2 = x b and (y — x) 2 — x* have cusps of the 
first species at the origin. 

23. Prove that x* — 2ax 2 y — axy 2 -j- a 2 y 2 — has a cusp of 
the second species at the origin. 

24. Prove that x* — ax 2 y — axy 2 -f a 2 y 2 = has a conjugate 
point at the origin. 

25. Prove that the multiple point of ay 2 = (x — a)'\x — b) 
at (a, 0) is (1) a conjugate point if a < b, (2) a double point if 
a > b, and (3) a cusp if # = b. 



CHAPTEE VIII. 

GENERAL DEPENDENT INTEGRATION. 

FUNDAMENTAL FORMULAS 

180. The differentials in the following twenty-two formulas 
are the fundamental integrable forms, to one of which we 
endeavor to reduce every differential that is to be integrated by 
the dependent process (Art. 51): 

/ v n+l 
v n dv = — — - + a 

/dv 
— = log (v) + C, or log v + log c = log cv. 

3. r<fdv = ^- + 0. 

J log a 

4. fe v dv = e v + C. 

5. / cos v dv = sin v -j- 0. 

6. / sin v dv = — cos v -f C. 

7. / sec 2 v dv = tan v + C. 

8. / cosec 2 v dv = — cot v + C 

9. / sec v tan v dv — sec v + C. 

160 



GENERAL DEPENDENT INTEGRATION. 161 

10. / cosec v cot v dv = — cosec v -\- 0. 

11. / tan v dv = log sec v -f C. 

12. J cot v dv = log sin v -\- O. 

13. / sec v dv = log (sec v -f- tan #) + (7. 

14. / cosec v dv — log (cosec v — cot v) + 6'. 

16. / %— — = cos" 1 1; + a 
^ fl - *; 2 






18 



:/ -*«*-. + <* 



19. /* / J! = sec" 1 v -f (7. 

•/ V 4/ v « _ i 

20. f- ^J?— = cosec- 1 w + 0. 
«/ t,lV _ i 

21. / > —£^ = r=:vers- 1 ?;+ (7. 



V2v -" 



y = log (v + iV ± m) + t a 



22. 

^ v* ± m 

In these formulas v may be the independent variable, or 
some function of this variable, and the process of integration 
consists largely in reducing or transforming any given differen- 
tial into one of the above forms. 



162 . DIFFERENTIAL AND INTEGRAL CALCULUS. 

REDUCTION AND INTEGRATION OF DIFFERENTIALS. 

181. Reduction of Differentials is the process of reducing 
them to integrable forms, and is effected chiefly (1) by constant 
multipliers, (2) by decomposing or separating them into their 
integrable parts, and (3) by substitution. 

182. Reduction of Differentials by Constant Multipliers. 
Principle. The value of any differential of the form c I dv 

remains unchanged if dv be multiplied and c be divided by the 
same constant. 

EXAMPLES. 
Find the following: 

_ C ndv . n . . v „ 

1. y = I — ■ — -. Ans. — - tan" 1 -=. + C. 

We reduce this to the form of 17, Art. 180; thus 




n C dv nVa / Wal 
y — - \ r = / ■ / — r 9 = Answer. 



/itdv v 

, (See 15, Art. 180.) Ans. n sin" 1 -j=. + C. 

Va - v 2 Va 



r f ' 

_ n T dv _ nVa f 



jv_V 
Va! 



= Answer. 



Let the student compare these results with formulas (13) 
and (15), page 65, and in a similar manner deduce formulas 
(14) and (16) on that page. 



GENERAL DEPENDENT INTEGRATION. 163 

_ 3 r (2x + 2)dx _ 3 / V(z 2 -f 2a; + 5) 
V ~ %J x* + 2x + 5~~2J x* + 2x-\-5 - An s. 

(Formula 2, Art. 180.) 

^ */2-9a; 2 V2 

r C %dx . _ , _, 

6. / - vers -1 3a; + (7. 



VQx — 9a; 2 

Mx 
2-f W |/14 """ ~ r 2 



r ?>dx 3 x , A /7 



8. /V^« i tan- 1 a; 2 +(7. 
^ 1 + x* 

r dx i tan _i ^ + a 

" t/ l-f-5o; 3 * ^5 

P dx 1 , ,- , _. 

-1. / . :. — sec" 1 aVf + 67. 

47 xV2x* - 3 ^3 

!. / ._ | vers * -tt+ C. 

;. f " ** , * cos" 1 a;^ + (7. 

^ V3 - 2a; 2 ^2 

1 fx^TQ' itan-^+a 



164 DIFFERENTIAL AND INTEGRAL CALCULUS. 

15. f ° dx T Vb sec- 1 xV* + a 

J xVZx* - 5 

16. f dx , i^sec- 1 ^ +C. 
J Vbx* - Zx* 

Some of the preceding examples may be conveniently solved 
by formulas (19) and (21), page 65. 



REDUCTION OF DIFFERENTIALS BY DECOMPOSITION. 

183. The process of reducing differentials to integrable 
forms consists largely in separating them into their integrable 
parts. 

184. Elementary Differentials. In these the necessary 
reductions are effected by the simple operations of algebra. 

EXAMPLES. 
Find the following: 

1. y ^tfo s- t log (4z 2 + 1) + I tan" 1 (2x) + 0. 

r{?>x -f5) _ r 'Sxdx r $dx 



+ 1 «/ 4ar + 1 ' t/ 1 + 4z a 



2 . f^J^Ldx. -(l-^^-Ssin-^ + a 

*/ |/l_z 2 

3. A ( 2 ~ 5 ^ _ <fc. 4- (^ ~ «*)* ~ 4^ vers" 1 a + £ 

^ V±x-2x' i V2 V2 

# L &' x Vx 2 — 1 J 



(a a - 1)* + cosec" 1 x + C 



GENERAL DEPENDENT INTEGRATION. 165 



/ Va + x d f Va + x _ a+ x 
Va-x X ' L Va-x VV — x 2 ' J 



Bin- 1 -- tV^^aj* + a 



a 



8 -/Vfl^ ?^+™* + log(l + *) + a 

185. Trigonometric Differentials. — In reducing these we 
use the elementary formulas of Trigonometry, such as 

sin 2 x -f- cos 3 % = 1, sin 2a; = 2 sin a; cos x, 
cos 2a; = cos 2 x — sin 2 a;, etc. 

EXAMPLES. 
Find: 

1. / sin 3 x dx. — cos x -\- -J cos 3 x -f C. 

sin 3 a; = sin x (1 — cos 2 a;) = sin a; — (cos 2 x) sin a?. 

/ (sin 3 x)dx = J sin x dx -j- / (cos a;) 2 ^(cos a;). 

2. I sin s xdx. = I (I — cos* x) 2 s'm x dx. 

— cos a; -f- f cos s a; — -^ cos 5 a; + C. 

3. / sin 7 x dx. — cos x -j- cos 3 a; — f cos 5 a; -j- 4" cos7 # + C- 

4. / cos 3 x dx. sin a; — ^ sin 3 x -f C. 

cos 3 a; = cos x(l — sin 2 a;). 



i 



166 DIFFERENTIAL AND INTEGRAL CALCULUS. 

5. / cos 5 x dx. sin x — § sin 3 x + \ sin 5 a; + C. 

In like manner / cos m x dx and / sin m x dx can be found 
where m is any odd positive integer. 

6. / sin 4 x cos 8 x dx. \ sin 5 x — \ sin 7 . a; -f- G. 

sin 4 x cos 3 a; = sin 4 x(l — sin 2 a;) cos x. 

In a similar manner / sin m x cos TC a?^a: may be found when 
either m or n is any odd positive integer. 

7. / sin 8 x cos* xdx. — \ cos 5 x -\~\ cos 7 $c + G. 

8. / sin 2 a; cos 7 x dx. -J sin 3 a; — f sin 6 a; -f- -fsin 7 x 

- \ sin 9 x + C. 

9. / sin 3 a; Vcos a; tfa;. — f (cos x)* -f -f (cos a;)* -f- G. 
10. / cos 2 a; dx. — + i sin 2a: -f- (7. 



i,/ 





cos 2 - a; = -J -f- -J cos 2a:. 


sin 2 a; dx. 


- — i sin 2a: -j- 0, 




or — — -J sin a; cos x-\- G. 


dx 


[=ft*x**-J logUnx+O. 


sin a: cos x 


* dx 


tan x — cot # + 01 


sin 2 a; cos 2 x 




1 sin 2 a; -f- cos 2 x 



GENERAL DEPENDENT INTEGRATION. 167 

, . /*sin 3 x dx i-i/i 

14. / 1 — . sec x 4- cos $ -4- C. 

«/ cos a; 

15. / sm . ^ Jav 1 tan 5 a; + i tan 3 a; -f 67. 
e/ cos 6 ^ 5 3 ' 

— tan* x sec 4 sc = tan 2 x(l -f- tan 2 a;) sec 2 a;. 



cos a; 
sin 2 a: 



cos x 

In like manner - — - — dx or - — — dx may be integrated when- 
cos x sin" a; 

ever m — n is even and negative. 



16. f-^ . tan x + i tan 3 a; + C. 

J cos 4 a; 

_ fcos-xdx , 

IT. / — — i . — i cot a; + (7. 

«/ sm a; 

186. Trigonometric differentials can often be more conven- 
iently integrated as indicated by the following solutions. 

18. / cos 4 x sin 3 x dx. — \ cos 5 x -\- \ cos 7 x -f- C 

Ay 



Make cos x = y\ then sin x = V'l — ^ 2 and dx 



Vl-y\ 

.-. ^cos 4 a; sin 3 xdx = J — y\l - %f)dy = - ±ij" + \y 7 + C. 

19. / tan 5 x dx. J tan 4 x — -J tan 2 a; + log (sec a:) -f- C. 

Make tan x = y: then Ja; = -=-= — -. 

,. f^xdx=f^dy=f[y^y^^dy 

= iy t -if + ^o g (f + i) + c. 

This method combined with that of Arts. 212 to 215 affords 
a complete solution of rational trigonometric differentials. 



168 DIFFERENTIAL AND INTEGRAL CALCULUS. 

187. Rational Fractions. A fraction whose terms involve 
only a finite number of positive and integral powers of the 
variable is called a Rational Fraction; as 

x* - 2x* - 13z 2 + 17 . 

ay = £-1-7; — r~K dx. 

* x 2 + 3x + 2 

To separate this fraction into its integrable parts, we first 
divide the numerator by the denominator and obtain 

dy = (* - 5x)dx + J% + ™ 2 dx. 

Again, by separating the fractional part of this quotient into 
two parts (its "partial" fractions), we obtain 

lOz + 17 . Idx , Mx 
-dx 



x 2 + 3x + 2 x + 1 ' x + 2 

, . a 7<fo 3d# 



tfa; 



+ 1 ' c/ z + 2 ' 
or 2 / = T - x + 71og(x+l) + 31og( 2; + 2) + £7. 

That is, y = . L0X + log (a + 1)> + 2) 3 + C. 



The first step in the above and similar operations is very 
simple, and it is our present purpose to show how the second 
step, the separation and integration of fractions whose denomi- 
nators contain a higher power of x than the numerator, may be 
effected ; and to render the process as simple as possible we shall 
apply it to particular examples in each of the four cases that 
may occur. 



GENERAL DEPENDENT INTEGRATION. 169 

188. Case I. When the simple factors of the denominator 
are real and unequal, 

EXAMPLES. 

7 (x + l)dx 

1. Integrate dy = ^ 3 + 6 ^ + 8 ^ 

The roots of z 3 + 6z 5 + 82 = 0, are 0, - 2, and -4; hence 
the factors of x 3 + 6z 2 + 8z are a;, x + 2, and a + 4. 

s + 1 -4 5 (7 ' 

Assume ^ + ^ + 8:g - - + ^2 + ^+4' ' ' (1) 

Clearing (1) of fractions, we have 

x + 1 = A(x + 2)(a> + 4) + i?(z)(z + 4) + C(s)(s + 2), (2) 
or x+l=(A + B+C)x* + (GA + 4B+2C)x+8A. 

Equating the coefficients of the like powers of x, we have 

A + B + C=0, 64 + 45+20=1, 84 = 1. 

Solving these equations, we find A — \ y B — \, and C = — f. 
Substituting these values in (1), we have 

x + 1 __1_ 1 3 



x 3 + 6z 2 + 8x 8x "*" 4(a + 2) S(x + 4)' 
dx , dx 3dx 



8z~ r 4(z + 2) 8(z+4)' 
, />c£e . . f* dx » f* dx 

= i log a + f log (as + 2) - f log (x + 4) + i log e. 



. l/cx(x + 2) a 



170 DIFFERENTIAL AND livTEGRAL CALCULUS. 

The values of A, B, and C may be obtained from (2), thus 



Making x = 0, we have 1 = 8 A 
Making x — — 2, we have — 1 = — 4i? 
Making x = — 4, we have — 3 = 80 



.: B= I 



Principle. In this case, to every factor of the denominator, as 

x — a, there corresponds a partial fraction of the form ■ . 

x a 

Find the following: 

3dx , (x — 2 \* 



c C Zdx , X — 7£ \* t n 

J x* — a 2 * T x + a 

*/5^TT log[(*-l)=(* + 2)V]. 

5 - Z ^+V-to - log ^— W+8)'+ G - 

6 . /*+^J fc £ + %. + 4, + tog^fe^-* + a 

J x — 4:X 3 2 & (a; + 2) 3 

189. Case II. 'TPften some of the simple factors of the de- 
nominator are real and equal, 

EXAMPLES. 

, ' _ , , , (x a + x)dx 

1. Integrate dy = (g , 2) , (g . iy 

Assume -, ^j-, rr = 7 ^ H „ + 



(x - 2)\x - 1) (x - 2)* * x - 2 * x- 1 
Clearing of fractions, we have 

z s + x = 4 (a - 1) + B(x - 2)(x - 1) + (7(3 - 2) 2 . 



GENERAL DEPENDENT INTEGRATION. 171 

Making x — 2, we have 6 = A; .-. A = 6. 
Making x = 1, we have 2 = C; .-. C = 2. 
Making a; = 0, we have = — A -f 2B -f 4C ; .*. 2? = — 1. 

a 2 + a 6 1 2 

•• (x-2)\x-l)~ (x-2f x-2 + ;~T 

^~~ (x-2y~x~^2 + x 



1' 



y V (3 _ 2 ) 2 e/ 3 - 2 + V s _ 1 

= ~ ^ZT2 ~ lo S (« - 2) + 2 log (x -1) + C 



= ^(^—27-5^+^ 

Principle. In this case, to every factor of the form (x — a) n 
there corresponds a series of n partial fractions of the form 

A B K 

(x - a) n ' (x - a)"- 1 ' " ' x — a 

Find the following: 

_ P(3x - l)dx . , , ox 8 , n 

r ^x 1 + 9z - 128)^ 
d - J (x - 3)\x + 1) • 

\-17log(x-Z)-8log{x + l) + a 



4 -/?+TO+i- i^s + ^C + D + a 
6 - J ^+W dx - 2WT7 + 3 los W+l + a 



172 DIFFERENTIAL AND INTEGRAL CALCULUS, 

190. Case III. Wlien some of the simple factors of the de- 
nominator are imaginary and unequal. 

EXAMPLES. 

1. Integrate dy = (g + ^*. + 4) . 

Here the two simple factors of x* -f 4 are x -f- 2 V — 1 and 
# — 2 4/ — 1 ; we may take these factors and proceed as in Case 
I, but the integrals obtained would involve the logarithms of 
imaginaries; to obviate this, we assume 

x __A_ Bx+ C 



(z + l)(z a + 4) x + 1 l z 2 + 4 * 
Clearing of fractions, we have 

x = A(x 2 + 4) + (Bx+ C)(x + 1). . . . (1) 
Differentiating (1), we have 

1 = 2 Ax + B(x + 1) + Bx + C. . . . . (2) 
In (1) making x — — 1, we have A — — \. 
In (1) making x = 0, we have C = — 4^4 = -f f. 
In (2) making x = 0, we have B = 1 — C — \. 

Principle. /w £te case, to every factor of the denominator 
of the form (x — ay -f 6 2 ^ere corresponds a partial fraction of 
Ax-\- B 



the form 



(x _ a y + y 



GENERAL DEPENDENT INTEGRATION. 173 

3 - /ftwt?' 4 tan_1 ■ - * tan_1 1 + a 

5 - f^fi d *- , + ilog^- l/3-tan-^ + C. 

6 . A«- 1 log ^±^|±_1 + tfj tan- f^ + a. 

191. Case IV. Wlien some of the simple factors of the 
denominator are imaginary and equal. 

EXAMPLE. 

1. Integrate dy = ^ + ^ _ ±y 

1 Ax + B , Cx + D , ^ 

Assume , . , D , 2 , -r- = , , , „ a + - , Q 4- 



(z 2 + 3) 2 (z - 1) (a; 2 + 3) 2 ' x* + 3 ' x - 1 
Clearing of fractions, we have 
1 = (Ax + 5)(«-l) + (Cx + Z))(z 2 + 3)(a-l)+.tf(a; 2 +3)". (1) 
Whence A = -\, B = - ±, C = - T V, 

1 »+_!_, 1 (a; + 1) , 1 tfqg 
«y - - 4 (a .. + 3) .«* 16 z 2 + 3 + 16 x - 1" 

-^7-in " ^^g (* 2 + 3) - -^--tan- 1 - 



* 8(z 2 + 3) 32 °^ ' ' 16 ^3 ^3 

+ ^i og(a; _ 1) _iy > _^_ 

The integration of differentials of the form -, — _ ' 1 . mt may 
5 {ax 2 -\- b) m J 

be more conveniently obtained by Art. 211 or 215. 



174 DIFFERENTIAL AND INTEGRAL CALCULUS. 

REDUCTION BY SUBSTITUTION. 

192. Irrational Differentials. To integrate an irrational 
differential which is not of one of the known integrable forms, 
we first rationalize it, and then proceed according to the previ- 
ous methods. 

To show, in a simple manner, how rationalization is to be 
effected, we shall apply the process to a few particular examples. 

EXAMPLES. 

Find the following: 

(2 Vz + i)dx 



' V J 2\ / x(x + 3Vx + 5)' 
Make x = z 1 ; /. dx = 2z dz. 

y = A%XT+6) = log v + 3z + 5 > = l0 « 1* + 3 ^ + 5 )- 

2 A> Vx + l)dx 

-n ~. tan" 1 Vx + C. 

2(x k + x*) T 

4. / =. [Make x = z\] sin -1 Vx -f- C. 

3 r # a r 1 — x* 

6. /%^-v [Makea = « 6 .] 

87 «;* + ** 

2 1/*- 3fz + 6f^- 61og(l + fe)-fC. 
/■ (s-l)efo |/^_ 2 



GENERAL DEPENDENT INTEGRATION. 175 

193. When a -f- to is the only part having a fractional 
exponent. 

Assume a -\- hx = z n , where n is the least common multiple 
of the denominators of all the fractional exponents; then the 
values of x, dx, and each of the surds, will be rational in terms 

of z. 

EXAMPLES. 
Find 



1. y = / x 3 Vl -f xdx. 



Assume 1 -+- x = z 2 ; then 4 1 + x = z. ... (1) 

Also x-z*-!, x* = (z°-iy, . (2) 

dx = 2z dz. . . (3) 
Multiplying (1), (2) and (3) together, we have 

fx % VT+xdx =f%z 2 (z 2 — l) 2 dz 

= ^ - K + \z* + C 



'-/ 



xdx 2(x — 2 )4^ 1 +x 



|/l + z 



4. y z(a -f- z) 4 efc. ^(0 + a)*(4z - 3a) + C. 



1 94. When y a -f 6a; + x * or ^« + to — # 3 is the only- 
surd involved. 



A differential containing no surd except Va -\-to-\- x 1 can 
be rationalized by assuming Va -\- ~bx-\- x 2 = z — x; and one 
containing no surd except Va -f- to — x 2 can be rationalized by 



176 DIFFERENTIAL AND INTEGRAL CALCULUS. 



assuming Va -f- bx — x u = (x — r)z, where r is one of the roots 
of a -j- bx — x 1 = 0. 

The process is illustrated in integrating the following im- 
portant differentials (see Ex. 30, 31, page 67). 

1. Find y = f- 



Va + bx + x* 

Assume Va + bx + x* = z — x; then 

- a 



a -j- bx = z* — 2zx, x = 



2z + b' 



2(z' + bz + a )dz 

ax ~ (2z + by ' (l > 



1 2z + b 



(2) 



Va + bx + x' z' + bz + a 

= log (2« + b) + a 

.: f- r—^r = log (2* + 6 +2 Va+bx+x') + 0. 

^ Va -\-bx -\-x 

When b = 0, 

2. Find y = ^ 



|/a -}- &z — as 2 



Represent the factors of a -j- bx — a; 8 = by # — r and 
— x, and assume 



|/# + bx — x* = V(x — r){r' — x) = (x — r)z, 



GENEBAL DEPENDENT INTEGRATION 177 

rz* + r' 



then t' — x = (x — ?-)z 2 , x 



z 2 + 1 ' 



dx - Hr-r')zdz - 1 _ *' + * / 9 v 

" (s a + l) 2 ' * * {l) Va + te _ ^ " (r' - r)s * ' ^ 



(1) X (2), f dX = = - 2 I 



\/a + l)x-x 2 J 1 + 2' 

- 2 tan- 1 2 -f C. 



... ^ ^ = - 2 tan- fA^-* + a 

When 5 = 0, ?• = + i/a, and r' = — Va, we have 



2 tan -1 \ / — — -! + C. 



Va - x' 1 V V 



V- 



a — x 



3. 



*/ a V2 + a; - a; 2 ^2 \ ^2 + 2x + V2 - a;/ * 



Assume V2 + a; - x 2 = V(2 - x)(l + a;) = (2 - z)z; then 

* = 7TT' ; & "' = ^p and i/2 + x - ¥ =7Ti' 

r dx _ r 2dz 1 . zV2 — l 

J x V2 + x -^ ~ J ~z 2 ~ 1 ~~ 1/2 ° g z | 7 2~ + 1 ' 

4 " Atf?5^i »t«-M* + •* + &=!) + a 

/» rgWa; t - 1 (a; + 3) 13 + 2? - x 2 

°- y wp^7" 3 sm ~i 2 • 



178 DIFFERENTIAL AND INTEGRAL CALCULUS. 

195. Binomial Differentials. Differentials of the form 

r 

x m {a + bx n )'sdx, 

where m, n, r, and s represent any positive or negative integers, 
are called binomial differentials. 

196. To determine the conditions under which a binomial 
differential is integrable. 

T 

I. When - is a positive integer the binomial factor can be 
s 

developed in a finite number of terms, and the differential exactly 

T 

integrated ; and when — is a negative integer the differential is 

a rational fraction whose integral can be obtained by the method 
of Art. 187, 212, 214, or 215. 



II. Assume a -f bx n = z s ; .*. (a -f bx n )s = z r , . . (1) 

i 

(z s — a^ 
x = 



p-f)> -=(-^r ■ • « 



and dx = ^rU^)- 



Multiplying (1), (2) and (3) together, we have 



m + 1 



z w (a + &&")• dx = ^z r + s - 1 (— ^ — J <&. . . (4) 

The second member of (4), and therefore the first, is in- 
tegrable when — - is a positive or negative integer, by Case I. 

III. Assume a + bx n = z s x n ; .\x n = a(z s - b)- 1 , 

1 1 m _ m 

x = a*(z s - b)~~~ n , x m = a»{z s — b) », .... (1) 

a + fc»" - ~^j, (a + ^ n )* = «'(** ~ &)~ ^dz y ... (2) 
1 z s — b 



GENERAL DEPENDENT INTEGRATION. 179 

S - --1 

and dx — a*z a -\z 8 — b) » " dz. (3) 

Multiplying (1), (2) and (3) together, we have 

r „ m + 1 r / m + 1 r \ 

z w (« + fe n )^r = a » + *(* s - b)~^ n * h r + s ~ x dz. (4) 

By Case I, the second member of (4) is integrable when 
j- — is a positive or negative integer. 

r 

Hence, x m (a -f- bx n )*dx can be integrated by rationalization: 

7Yh I 1 

I. WJien is an integer or 0, by assuming a -\- bx n = z s . 

XT - _ m 4- 1 , r . . , _ , 

11. It ^ew 1 — is an integer or 0, by assuming 

a -f £>£ n = z s x n . 

"When the differential reduces to a rational fraction, which is 

i m + 1 , -. • , • m + 1 r ' ■ „ 
the case when + 1 1S a negative, or ! 1 f-1 aposi- 

n n s 

tive, integer, it is less laborious to integrate by a method to be 
subsequently given. 



1. Find A 5 (l + a 8 )* 



EXAMPLES. 
dx. 



TT m + 1 5 + 1 . 

Here — ■ = — - — = 3, an integer, and s = 2 ; hence we 

assume 

1 + Z 2 = 2 2 ; .-. (1 +*■)* = *, . . . . (1) 

x* = z* - 1, z 6 = (^ + l) 3 (2) 

Differentiating (2), 

6x b dx = 6(^ 2 + l) 2 z af».-. ■ (3) 



180 DIFFERENTIAL AND INTEGRAL CALCULUS. 
Multiplying (1) and (3) and dividing by 6, we have 

/V(l + z')*dx - yV + lyMz 

= J{z' + 2z> + z')dz 

= K + K + V+o 

= {(I + *•)* + 1(1 + *•)* + J(l + «.)•+ ft 
Find: 

2. yva + *•>*€&. (i + x>) i (^~^\ + c. 

3. /V(l + z'fdz. 

A( l + ^)V _ | (1 + j,)! + A(1 + ^i + a 

4. yV(l + zY'dz. (1 + ^ , )*(^f : - 2 -) + 0. 

t>. fz-M + zr'dz. -5S=i + a 



Vl + 



uri + r 



Here 



m + l_-2 + l__l. m + 1 r__l 3__ 

~1^-— ~2 - 2' and n +s~ 2~2~ 2 ' 

an integer; hence we assume 

1 + *■ = zV; .-. (1 + x*)- % = (Z ' 7, 1} ; •.. (1) 



■ = <-l; . . . . (2) 



-77^—. • • (3) 



OENEBAL DEPENDENT INTEGRATION. 181 

Multiplying (1), (2), and (3) together, we have 

= -{* + -) + c, 

where z = -Vl 4- x 2 . 

x 

6. / '(1 + af)~*dz. [ro = 0.1 —JL= + a 
J Vl + x 2 

7. fx'il - 2x T ^ - (1 + ^)(1 - 2x^ + c 

8. fx'Ha + xT*dx. — + 2a . + a 

INTEGRATION BY PARTS. 
197. Integrating both members of 

d(uv) = udv + ttf?M 
and transposing, we have 

/ udv = uv — I vdu, (A) 

which is the formula for integration by parts. It reduces the 
integration of udv to that of vdu, and by its application many 
differentials can be reduced to one of the elementary forms. 



EXAMPLES. 



1. Find / x* log xdx. 

Assume u — log x; then 

dv = x 2 dx, du = — -, and v = / x' 
x <J 



dx = -. 



182 DIFFERENTIAL AND INTEGRAL CALCULUS. 

Substituting in (A), we have 

/x\ fx 2 dx x s log x x 9 „ 
x* log xdx = - log x -J -y- = — -£■ -q +C. 

2. Find I sin -1 x dx. 

Assume u = sin -1 x: then rfv = dx, v = x, and du = , 

Vl-a; 2 

Substituting in (A), we have 

/sin"; 1 # £?£ — a sin -1 x — / — = a; sin -1 » + (1 — x^A- C. 

Find the following: 

3. / tan -1 x dx. x tan -1 x — log (1 + x*) k -f- C. 

4. / as cos a; fc a: sin a; + cos a; + & 

5. fxe ax dx. e ax (^ - ^\ + 0. 

Make dv = e ax dx: .: v = — , u = x. 

a 

Sometimes two or more applications of the formula are re- 
quired, as in the next example. 



6. 



fx*e°*dx. e ax [~ - - 9 + 41 + G. 

J \_a a * ^ a 3 J ' 



n ax 



Make dv = e ax dx: .'. v — — , u = x*, du = 2x dx. 



a 



I* » n~i e ax x 2 n Pe ax xdx 
/ x*e ax dx = 2 / . 



Now apply the formula to the last term, as in Ex. 5, and we 
obtain the entire integral. 



GENERAL DEPENDENT INTEGRATION. 183 



7, 

Va? - x* 



xdx 



Make dv = ; .*. v = — |/a 3 — x% and w = x\ 

Va? - x* 

r_ ^dx_ = _ ^- r — 2 A^ _ x y dXu 

J Vcf-x* J 

In a similar manner we may integrate any binomial differen- 
tial, or by continued application of the formula reduce it to a 
simpler form, but by the method of Art. 211 the result may in 
general be obtained with less labor. 

8. / x s cos x dx. 

Make dv = cos xdx; then v = sin x, n — x s , du = 3x*dx. 

I x 3 cos x dx = x 2 sin x — J 3ic 2 sin x dx. 
Again, make dv=- sin x dx; then v = cos x, u — 3x 2 ,du = 6xdx. 

— I 3x 2 sin x dx = dx 2 cos a; — / 6a: cos a; dfa:. 
Again, make dv= —cos a; dx, then a = — sin x, u = 6x, du = 6dx. 
.'. — / 6x cos xdx = — 6a: sin x--{- / Q sin a; ^a( = — 6 cos a:). 

/. / x s cos xdx = x 3 sin x-{-3x 2 cos a;— 6a; sin a;— 6 cos »+ (7. 

9. / log x dx. x(\og x — 1) + C. 

/x 4 X* 

x s log x dx. — log x — — -f 0. 

11. yV log 2 x dx. ^(log 2 x - f log a; + f ) + (7. 



184 DIFFERENTIAL AND INTEGRAL CALCULUS. 

12. Je x x"dx. 0*(£*-4z 3 +12z a -24z+24) + a 

. / xa x dx. =-^— [x — z J + C. 

e/ log a \ log aJ 



13 



log a \ log 

14. fx*&~ x dx. — e~ x (x 2 + 2x + 2) -f (7. 

15. A; V*£fc. (a' - — + -f - ~ )— + a 
J \ a a a / a l 

16. yV sin" 1 a; <fo. |- sin" 1 x + ^pVf^T 2 -f (7. 

18. yV (log z) 2 fe j [(log z) 2 - * log 3 + £| + (7. 

y (a 2 - z 2 ) -V*& - I (a 2 - z') } + y sin" 1 ?- + C. 

f{a* + tf)-*sfdx. - (a 2 +z 2 )* - ^log (« + rf*+&)+0. 



19 

20 



REDUCTION FORMULAS. 

198. Reduction formulas are formulas by which the integral 
of a differential may be made to depend on the integral of a 
similar, but simpler, differential. 

199. To find the reduction formula for / x p (log x) n dx 9 
where n is a positive integer. 

Assume dv = x p dx and u = (logo;)"; 

x p+1 dx 

then v = — r^r and e?w = nClosx)*- 1 —. 

p + 1 » 

Substituting in (A), Art. 197, we have 
/a»(log *)"«fa = ^(y _ _2_yV ( i og V y- H x, (1) 



GENERAL DEPENDENT INTEGRATION. 185 

in which the proposed integral depends upon another of the 
same form, but having the exponent of log x less by one. By 
successive applications of this formula the exponent of log x 
is reduced to zero, and the proposed integral is made to depend 

upon the known form / x p dx. 

Cor. I. If the given integral were ) X (log x) n dx, where X 
is any algebraic function of x, we should have 

J X(logx) n dx = X^logx) 71 — nj —(log x) n - x dx, 
where X t = I Xdx. 

1. J x z log 3 x i 



EXAMPLES. 
dx. 



Here p = 3, p + 1 = 4, n = 3 and n — 1 = 2. Substituting 
in (1), we have 

/ x 3 log 3 xdx — \x* log 3 x — | / x 3 (log xydx. 



By applying the formula to the last term, etc., we obtain 



log x--log*x + ~\ogx jjp- 



+ 0. 



J x' log' X = j 

2. fx' (log x)' dx. J [(log x)' - i log x + T Vj + C. 

3 -/tr|- ^logx-ioga + ^ + a 

x= x l 



(i + z) i i + x 



186 DIFFERENTIAL AND INTEGRAL CALCULU.3. 

200. To find the reduction formula for / a x x n dx, where 
n is a positive integer. 

Let dv — a x dx and u = x n ; 

a x 

then v = ; and du = nx n ~ x dx. 

log a 

.-. Art. 197, fa x x n dx = f^_ _ _»_ f a * x n-i dXt _ _ m 
v log a log a*J 

EXAMPLES. 

r 37 a* r 3 3^ 2 , 6^ 6 ~i , „ 

1. / a x x*dx. r iC — ; — - -f ; — 5 , — — + C. 

J log a L log « log « log rtj 

a x x 3 dx = . — : / a x x 2 dx. 

log « log ««/ 

By further applications of (1) we obtain the desired result. 

2. / a x x l dx. 



log a 



ix 9 , 12z 2 24a: , 24 



+ G 



log a log 2 a log 3 « log 4 a 

t/ ft L « a a J 

201. To find the reduction formula for / x n cos ax dx 
and / x n sin a» ate, where n is a positive integer. 

Make u — x n and dv — cos axdx; 

sin tfo; 



then c??/ = nx n ~ 1 dx and v = 



a 



/x n cos <7# 6/.T = — x n sin a# / x 11 ' 1 sin «# cfo. 
a a<J 



GENERAL DEPENDENT INTEGRATION. 187 

Similarly, we find 

/x n sin axdx = x n cos ax A — / x n ~ l cos ax dx. 
a ad 

Hence, in either case, the integral can be made to depend on 
the known form / cos ax dx or / sin ax dx. 

EXAMPLES. 

1. / x s cos x dx. x % sin x -f- dx 2 cos x — 6x sin x — 6 cos x -j- (7. 

2. / a; 4 sin # dx. 

— .r 4 cos a; -f- 4:X S sin a; -{- 12# 2 cos a: — 24a; sin x — 24 cos # + C- 

202. To find the reduction formula for Xsm~ l xdx, 
JTtan -1 x dx, etc., where X is an algebraic function of x. 

Make u = sin -1 x and dv = Xdx; 

du = —j=^-z—^ and v = I Xdx = X x (say). 



then 

Vl—x 



I Xsin l xdx = X 1 sin -1 x — I 



X x dx 



EXAMPLES. 
x 3 tan- 1 x x 2 log (1 + x 2 ) 



1. J x 2 tan" 1 z *fc. — + 6 J + ft 

x 3 
Here X=£ 2 and X 2 = — . 

2. f ^^* dx . x tan- 1 s- Ktan" 1 z) 2 - } log (1 + af) + (7. 

3. J x 2 sec~ 1 xdx. i£ 3 sec _1 £— 1(£ 2 — 1)*£— -J-log (x-\-Vx 2 — \-\-C. 

Reduction formulas for binomial differentials are deduced in 
Art. 215. 



188 DIFFERENTIAL AND INTEGRAL CALCULUS. 

203. To integrate . 

a -j- o cos x 

dx dx 



a-\-bcosx I . 2 x . a x\ j 2 x . a x\ 
alcos* - + sin 2 - J -f- 5(cos 2 - - sm 2 -J 



dx 



T X 

{a + b) cos 2 *- + (a — I) sin 2 - 



sec 2 —dx 



(a + I) + (a - I) tan 2 x 



■■■/-. 



dx 






= 2 



H) 



4- 1 cos x I , , ,x , / za i 2 a* 

(a + 6) + (a — 5) tan - 

which is readily reduced to the form 

fprrrp or i/Vtry acoordin s » a > or < b - 



dx 



/dx 
— — — -. — ■ can be found. 
a + b sm x 

* * m m • a d% ., dx 
204. To integrate and . 

sm x cos x 



/dx r idx _ Asec 2 \x jdx m 

sin x ~ J sin \x cos \x J tan \x * 

I 4^— — log tan \x = log y - 
•/ sm x . 1 



cos a; 



sin a; & . 1 + cos a; 



/" dx r dx 

Agam ' J^ = JMf^) 

= -log tan (~ _|j-f & 



GENERAL DEPENDENT INTEGRATION 189 



APPROXIMATE INTEGRATION. 

205. The number of differentials which can be integrated 
xactly is comparatively very small, yet the approximate value 
of the integral of any differential may be found when the dif- 
ferential can be developed into a convergent infinite series each 
of whose terms is integrable. This is the last resort in separat- 
ing a differential into its integrable parts. 

EXAMPLES. 
Find the approximate integral of the following: 

: ., dx x x 2 x 3 x* , , 

Expanding ——- by division, we have 

1 1 x , x 2 x 3 ■ , 



a 4- x a a 2 a a 



y = f ( ^- x a^i-¥ +&to ) dx - 



o j dx 

By division, 



1 + x 2 



o j dx 

3. ail = — . y = x 

J VI + x 2 
By the Binomial Theorem, 

1 





y 


= x - 


x 3 x h 

~ 3~ + 5~ 


x 7 

7 


+ etc. 


1 - 


- X 2 + X* 


-x« 


+ etc. 








x ' + 

2.3 ' \ 


dx 6 
2.4. 


3. 


ox 7 


+ etc. 




5 2.4 


.6.7 


irei 
= 1 


a, 

_?_! + 


3x* 
2.4 


— etc. 







Vl + x 2 

4. dy = x i (l- x'fdx. y = %x%- \z* - &x* - etc. 

Develop (1 — x 2 f, multiply by x*dx, etc. 



190 DIFFERENTIAL AND INTEGRAL CALCULUS. 

5. dy = z 3 (cos x)dx. y = ~ - ^ + ~ - etc. 

/I ic 2 3# 4 \ 

6. dy = x> sin- 1 xdx. y = x 1 ^- + — -f — -f etc. J. 

206. Development of Functions by Exact and Approxi- 
mate Integration. Two or three examples will suffice to illus- 
trate the process. 

EXAMPLES. 

/dx 
— — = log (a + x), exactly (omitting C); 

/dx x x 2 x 5 
^ + i = a ~ *? + 3? ~ et °" a PP roximatel y- 

x x 1 X* 

.'. log (a + x) = — + --;; — etc. 

a 2a 2 3a 

/^ ■ 

- _ = log (» 4- Vl 4- x 2 ), exactly: 

Vl+x 2 , 

r dx x* 3x b . , _ 

/ — = x 4- — etc., approximately. 

«/ Vl + x 2 6 ' 40 V^ J 



,-. log (x + Vl + x 2 ) = x--+~ - etc. 



3. / - — : — - = tan -1 x, exactly; 

t/ J. —r— X 



dx 



/ax x ., x x' , . , _ 

j-— v = a? .— y + -g ^ + etc., approximately. 

.t 3 £ 6 # 7 
.'. tan -1 x ~ x — V + "^ tt + etc. 



3 ' 5 



CHAPTER IX. 
INTEGRATION— (Continued). 

INDEPENDENT INTEGRATION. 

207. Increments Deduced from Differentials. We have 
seen that the increment of a function is the sum of the differ- 
ential and the acceleration; hence, when the former is known, 
we can find the differential by simply removing the acceleration. 
Taylor's formula enables us to reverse this operation in many 
cases, and find the increment when the differential is known. 

Let u=J(x). 

Increasing x by It, we have, by Taylor's formula, 

u + Au = f(x + h) 



n 



Au=f(x + h)-f(x) 

2 ' ' " J yuf \n 



=/'(*)* +/"(*)? + • • • / n w4r • • - ( A ) 



in which du =f'(x)h. 

Therefore, when the differential of a function of x is known, 
the increment may be found by taking the successive derivatives 
of the differential coefficient, and substituting them in (A). 

When f n (x) = 0, and each of the subsequent derivatives of 
f(x) = 0, the series will be finite and express the exact value of 
Air, otherwise the series will be infinite, and, if convergent, will 
give the approximate value of Au. 

191 



192 DIFFERENTIAL AND INTEGRAL CALCULUS. 

EXAMPLES. 

1. If du — (x* — 5x + 6)dx, what is the value of Au? 
Here /'(se) = a 2 — 5x -f- 6. Differentiating this, we obtain 

/"(a) = 2x - 5, /"'(a) = 2, / iv (z) = 0. Substituting these 
values in (A), we have 

Au = (x 2 - 5x + 6)h + (2a; - 5)^ + ~. 

2. If du = (x 2 — 3x — 10)h, what is the value of Au ? 

Au = (x* - Sx — 10)h + (2x - 3)£ + | 3 . 

3. If du = (x 3 — 7x 2 -j- 12x)dx, what is the value of au ? 

au = (3x* - Ux + 12)^-+ (3a - 7)^ + h * ■ 

4. Find aw when du = sin a fc 

h* . h 3 h* 

au — cos x - sin x — — cos a; — + etc. 

5. If du — (Vl -\-x)dx, by how much will u be increased 
when x is increased by h ? 

z/w = (Vl + a;)A+ (1 + x )~^~ (1 + a)"* — + etc. 

6. Find aw when *7w = log x dx. 

h> h 3 h* h> 

au = ^-W + i^~2-ox-* + etG ' 

7. A function is increasing at the rate of 4:X 3 dx; find its suc- 
ceeding increment. Au = 4x 3 h + 6x*h 2 + ixh 3 + ^ 4 » 

8. At the end of ^ seconds the velocity of a body is 

ds 

— = (3£ 2 — 2#) ft. per second; 

find the distance it will travel the following second, dt being the 
unit of time. As = (3f - 2t)dt + (3* - l)df + aT. 

9. The rate of acceleration of the velocity of a body is 

dv 

— - == (St -f- 4) ft. per second; 



INTEGRATION. 193 

find the increment (1) of the velocity (v), and (2) of the distance 
(s) for the following second. 

Av = (3t + 4)dt + \dt\ v = y (3* + 4)<7tf = ff + 4* + (7. 

Js = (ff + 4* + 0)<« + (it + 2)tff + \dt\ 

208. Increments as Definite Integrals. In Fig. 5, where 
u = area of OBPA, Au — the area of BCP'P, which is evi- 
dently the integral of du between the limits x and x -\- h. In 
general 



%J X 



>x+h 

f'(x)dx=f(x + h)-f(x). 



For, since df(x) -f'(x)dx, ff'{x)dx =f(x) + G. 

/*x+h — _ x+h 

I f'(x)dx = \f(x) + C\ =f(x + h) -f(x). 

*J x * 

Therefore (A) may be written 

r x+h h* h 3 h n 

J f'{x)dx =f'(x)h+f"(x)^ + f'"{x)j . . ./%e)j-...(B) 

By this formula we can obtain exactly, or in the form of an 
infinite series, the definite integral of any function of a single 
variable, and the operation does not involve the reversing of any 
of the formulas for differentiating. But, in general, this method 
is much inferior to that of dependent integration, since by the 
latter many differentials can be integrated in finite terms which 
by the former could be expressed only in the form of an infinite 
series. However, it forms an important part of the theory of 
differentials and integrals, and is often useful as a method of 
approximation. 

More convenient formulas for practical purposes will be 
derived from (A), but before doing so let us apply (B) to the 
following illustrative examples. 



194 DIFFERENTIAL AND INTEGRAL CALCULUS. 



1. Find the area BCP'P, the equation of APP f being 

A 




y = x 2 — lx + 12, where x = OB, y = BP, and BC = h or dx. 
Let w = area of OB PA, then dw = ydx. 

y=f'(x)=x 2 -7x + 12, f"(x) = 2x-7, 

f'"(x) = 2, /*(*) = 0. 

Substituting in (B), we have 

' x+h h* h 5 

(y)dx = (x* - Ix + 12)^ + {2x - 7)y + y = area of BCP'P. 



I 



Cok. I. To find the area of OEA we make x — and h = 
OE — 3, and obtain 13J. To find the area of EnF, we make 
x = 0^ = 3 and ft = j£F = 1, and get — f 

Let the student solve the following in a similar manner. 

2. The equation of a curve is y = x* — '6a?* -f- Ha? — 6; find 
the areas of the two sections enclosed by the curve and the axis 
of x. i; — i. 

209. A More Convenient Series. In (A), by makiug 
x = 0, then making h — x, and writing / f'{x)dx for f{x), 
we have 
//'<*)** =/(0) +f'(0)x +f"(0)~+ . . -f"(0)y . . (0) 

This formula may be obtained by developing f'{x) by 
Maclaurin's formula, multiplying by dx, and integrating each 



INTEGRATION. 195 

term separately, but as we are now exemplifying the method of 
independent integration, we will apply (C) directly to one or 
two examples. 



1. Find f X {Zx* - 14a + 5)dx. 



Here f'{x) = 3z 8 - 14a; + 5, .-. /'(0) = 5; 

f"(x) = 6x- 14, .-. f"(0) = - 14; 

f'"(x) = 6, /./'"(0) = 6. 

Substituting in (C), we have 



J^hx* - Ux + b)dx = x 3 


- W + 5x. 


2, J X {x % - 6x 2 + !)dx. 


± x * _ 2x 3 + 7a?. 


3. f*(3z A - 2x* + x*)dx. 


|^ _ ^ + i^. 



210. Bernouilli's Series. In formula (B), by making 

/x nx 

f'(x)dx = — I f'(x)dx, we have 

£ f'(x)dx =f'(x)x -f"(x)^- +...-(- l)"f"Wy • (D) 

This formula, called Bernouilli^s Series, like formulas (B) 
and (C), shows the possibility of expressing the integral of 
every function of a single variable, in terms of that variable, 
since the successive derivatives /"(#), /'"(#), etc., can always 
be deduced from f'{x). Hence, in all cases where the series 
are finite or infinite and convergent the integral may be ob- 
tained exactly or approximately. 

In finding ff'{x)dx by (B), (0), or (D) the limits of the 

difference between the approximate value found and true value 
may be determined as in A 6 of the Appendix. 



196 DIFFERENTIAL AND INTEGRAL CALCULUS. 



INTEGRATION BY INDETERMINATE COEFFICIENTS. 

211. The process of integrating binomial and trigonometric 
differentials by successive reductions is generally very tedious, 
and it is our purpose now to present a method which is gener- 
ally less laborious, and which is also applicable to many other 
classes of differentials. 

Let u r vdx be the differential to be integrated, where u and v 
are functions of x, and let us assume 

CvTvdx — u r+1 f(x) + kfu s v x dx, . . . . (E) 

in which it is required to find the function f(x) and the con- 
stant k. 

In examples where / u r vdx can be expressed under the 

form of u r+1 f(x), we shall find k = 0; and when this is not the 

case, / u r vdx will be determined in terms of / u s v^dx, which 

we can generally make of a more elementary character by as- 
signing suitable values to s and v % . 

Another advantage of expressing the required integral under 
the form of (E) arises from the fact that u r+1 /(x) often vanishes 
for the desired limits of integration, in which case the definite 
integral depends on the last term only. 

Differentiating (E) and dividing by u r dx, we have 

v = (r + l)fjjM+ «/*(*) + *»,«•-.. . . (F) 

The simplest and easiest method of solving this equation for 
f(x) and k is by indeterminate coefficients, as illustrated in the 
following examples. 

212. Case I. When k = 0; — Independent Integration. 

EXAMPLES. 
1. Find y (1 + x")~ k x\lx. 



INTEGRATION. 197 

Comparing this with (E) we have u — 1 -f- x*, r = — -J, 

j— = 2x, and v = x\ Substituting in (F), we have 
ax 

* = &z) + (l + aW{*) + && + *)"*- . ■ (1) 

The first member of this equation being of the fifth degree 
the second must be also; hence, f(x) must be of the fourth de- 
gree, and since u involves only the second power of x, we may 
assume 

f(x) = Ax* + Bx 2 + (7, .\ f\x) = ±Ax 5 + 2Bx. 

Substituting in (1), and arranging in reference to x, we 
have 

x* = bAx b + (3£ + ±A)x* + (C + 2B)x + kv^l + x*) s+ K 
1 = 5A, 3£ = - 4.4, C = - 2B, Jc = 0, 
or A = l B = -&, C=&. 

These values determine f(x), which substituted in (E) gives 
f(l + xY'v'd* = (1 + * ! )*[-K - ^ + &*] + ft 

2. Find yvu + *•/**. - (i + «■>• (i.. - r y + a 

Here u = 1 + a", r = -J, -7— = 2z, and «; = a" 6 ; .•. (F) gives 

*-* = ft^(a) + (I + «V(*) + *» I (l + ^)- 1 : . . (1) 

In order that the two members may be of the same degree 
we assume 

f(x) = Ax' 5 + Bx~ 3 + Cx- 1 ; .-. f'(x) = - 5Ax-«-3Bx- 4 - Cx~\ 

Substituting in (1), and arranging in reference to x, we find 
A = — \ 9 B = T \, G — 0, h = 0; this determines f(x); which 
substituted in (E) gives the desired result. 



198 DIFFERENTIAL AND INTEGRAL CALCULUS. 

A careful inspection of the previous examples suggests the 
following rule for determining the form of f{x) for binomial 
differentials of the form (a -j- bx n ) p x m dx, v being x m : 

I. When m is positive the highest exponent of ' x infix) will 
be m — n -f- 1. 

II. When m is negative the algebraically lowest exponent of 
x in fix) will be -$ m -\- 1. 

III. The remaining exponents decrease or increase alge- 
braically by n. 

The rule is also applicable when u is a polynomial in which 
n is the highest exponent of x, provided that the exponents of x 
in f(x) increase or decrease by the least difference between the 
exponents of x in u. 

When r is a fraction and u a polynomial of a higher degree 
than the second, the differential cannot ordinarily be inte- 
grated; or, more accurately, its integral cannot ordinarily be 
finitely expressed in terms of the functions with which we are 
familiar. The exceptional or integrable cases are, in general, 
where u, v, and r are such that it is possible forf(x) to have as 
many coefficients as there will be independent equations between 

the coefficients in equation (F), and where k I v x u s dx is or 

one of the integrable forms. In a differential of any given form 
the conditions of integrability may often be determined by the 
present method. 



r dx Vx 2 + Qx -f 15 f 1 1_ 1_\ _! q 



Vx* + 6x + 15 ( 1 


1 


-45 ^ 3 


2x* 


15, r = — i, and v 


= x-'- 



Vx % + 6x + 15 

Here u — x 2 -j- § x + 15, r = — i, and v = x~^\ hence the 
lowest exponent of x in fix) will be — 4 + 1 = — 3, and the 
others will increase by 1, giving fx = Ax~ 3 -J- Bx~ 2 -f- Ox~ x -\- D. 

The process can also be applied to many differentials in 
which v is a polynomial, as in the next example. 

r 3z 2 -f bx -f 5 , : /q~ i I \ . 



INTEGRATION. 199 

Here u = x* + 2x -f 3, r = — i, y = 3a; 2 -|- 5# -f- 5, and. /(a;) 
is of the form Ax -j- B. 



. r ax ^_ri^,i a . 



+ C. 



Here # = 1 -f- x*, r = — %, v = 1; and making v 1 = 1, s = 
- h (F) gives 

1 = - 53/0*) + (1 + x*)f'\x) + k(l + z 2 ) 3 , 

where /(a?) is evidently of the form Ax 5 -f ito 3 + (7a;. 

213. We have seen (Art. 185) that sm m x cos n x dx can be 
easily reduced to an integrable form when either m or n, or 
both, are positive odd integers, or when m -f- n is an even integer 
and negative. In such cases, m and n being integers, the inte- 
gration may be effected by the independent method, as in the 
two following examples; but this method of integrating such 
differentials is introduced and recommended chiefly for its bear- 
ing on the cases in which the above conditions do not exist, and 
which are usually solved by successive reductions. 

C • s 4 7 5 !~ s i n4 x , 4 sin 2 x , 8 ~| , „ 

6. / sin x cos x dx. — cos x — —r- 4- — — — -f — — -f- C. 
U |_ 9 63 oloj 

We may make u = cos x, r = 4; then -r~ — — sin x and 
v — sin 5 x. Substituting in (F), we have . 

sin 5 x = — 5 sin xf(x) -j- cos xf'{x) -{- Tcv x cos 3 " 4 x, 

where f(x) is evidently of the form A sin 4 x 4- B sin 2 x -\- C, 
and hence /'(a;) = (4J. sin 3 x + 2i? sin a;) cos x. 

.*. sin 5 a; — — 5 J. sin 5 sc — 5B sin 3 a; — 5C sin a: 

-f- (4J. sin 3 x -f 2i? sin x) cos 2 a; -f- etc. 

'Now, substituting 1 — sin 2 x for cos 2 a:, reducing, and arrang- 
ing with respect to sin x, we have 



200 DIFFERENTIAL AND INTEGRAL CALCULUS. 

(1 + 9 A) sin 5 x + (IB - ±A) sin 3 x + (5 - 2B) sin x 

-f Tcv x cos s_i x = 0. 

.-. 4=--i, B=-^ #=-1** * = 0. 



C dx sin a; f 8 . . 4 . . . ., H . ~ 

7. / — s— . — r— — sm 4 a; — - sin 5 x-\-l \ + C. 

J cos 6 x cos 5 # [_ 15 3 J 

Make w = cos a\ ?• = — 6, -=— = — sin #, v = 1, 5 = 0, 

ax 

v l = 1, and we have from (F) 

1 = 5 sin xf(x) -j- cos %f'(x) + h cos 6 a?. 
Assume f(x) = ^4 sin 5 # -f ■# si* 1 * ^ + Csin x. 

.'. f'(x) = bA sin 4 x cos # -f- 3i? sin 2 a cos a; + C cos a. 
Substitute, reduce, etc., and we find 

A = ~-, B=-~, G = 1, k = 0. 

This method of integration can often be applied to other 
classes of differentials, as in the next two examples. 

/x h ( 2 9 ■ \ 

x* log 2 x dx. - (tog 2 x --logx + g 5 J + <?• 

Make w = sc, r = 4, w = log 2 a;, and assume 
f(x) = A log 2 x -f B log x -\- C. 

9. /W& *■« (- - ?? + 5? - 4-) + a 

J \ a a? ' « 3 a* } 

du 
Make m = e ax , r = 0, s = 1; then v = e ax x\ and -=-7 = «e ax ; 

substituting in (F) and dividing by e ax , we have 

where we evidently have f(x) = Ax 2 -f Bx 7 -j- Cx -\- D. 



INTEGRATION. 201 

Find: 

r x a dx , /i 2\ 

10 -Jw^' -^-^V 2 + 3) + ^ 

11 C x * dx ,/-. A 4 , 1.4 , . 1.2.4N 

n - 7 yf=7" - * "*V +375^ +E1T5) + a 

yi ^W* + 5'.7* + 3.5.7^ +1.3.5.7J+ ' 
13. J x-\l + a 2 pVa. — ^ (- + 2a) + 0. 



Assume f(x) = (- Ok 

a 



Vjl /• a'ds ___/ 4aa 2 , 8a 3 \ 1 , _ 

14. / — -. Vrt + ^ 2 3a 4 _-+____ + CI 

«/ |/« + ^ 2 \ b l b 2 Jlbb 

15 /• ?*_ s/2te"_ 1 \ • 

' ■ t/ ( a _j_ ja*)l' (a + 6a 2 ) f \ 3a 2 "^ a) "*" "' 

16b / (^+^+»)fe vr+i^+D + a 

^ r 1 + * 

-.« /* B /o , o .\*7 (2 + 3a: , ) l /9 ■ 24 „ , 32 \ , „ 

17. y x\2 + 3a 2 )^a. 1-L_ ^_ a* _ _ a* + _ j + £ 

18. /ar^l - 2a 2 )"^a. - (1 - 2a 2 )*(-^^) + (7. 
g 4 — 2x + £)#3 

tV ~ 4X + 1 

(a 3 + a 2 ) da 

V5| + 4a 4- a 2 ' 



,« /"(a 4 — 2a 4- £Wa /~\ 

e/ l/fU \A~\rf V 5 ' \ 3 6 10/ 



«-, A 2 £ 5 - 5 ^ 4 - 6a 4- 7Wa 

21 - •/ K-5* + 2)» • ( *° " 5 * + 2)5(X - 5) + °- 

22. / sin 6 a cos 8 a dx. -J sin 6 a(cos 2 a 4- J) 4- CI 

Make u = sin a, r = 5, v = cos 3 a. 



202 DIFFERENTIAL AND INTEGRAL CALCULUS. 

23. f'—^dx. - —L- Tsiii 4 x + 4 sin 2 a _ 8~| + C 

«/ COS 2C 3 COS £ |_ J 

/^ 6 r 1-1 1 ~| 

x 5 log 3 a<fo. - log 3 x — - log 2 a + - log z - — + (7. 

/ e ax x*dx. 



25. / e^x"ax. 
log 2 # 



re 4 4a; 3 4. 3^ 4 .3.2a; 4.3.2.1 "! 



26 



f 



dx. 



2z 2 



log 2 x -j- log a; + 



+ c 



214. Case II. When k is not = 0; — Dependent Integra- 
tion. 

EXAMPLES. 
1. fx\\-x^dx. (l-^^-g-^+^sin-^ + a 

The differential # 4 (1 — x^fdx has the same binomial factor 
as (1 — x^^dx, whose integral is sin -1 x; hence, by expressing 
the former in terms of the latter, the required integral may be 

obtained in terms of / (1 — x*)~^dx. Thus, 

x\l - a-)* = x\l - s»)(l - x*)-* = {x'-x'Hl-x'y*. 

du 
* .% u =1 — x*, r — — i, — = — 2x, v = x* — x% which sub- 
stituted in (F) gives (making s = r and v x = 1) 

x * - x * = _ X f(x) + (l _ xY'{x) + k, 

where f(x) — Ax b -f- Bx % + Cx; and proceeding as before, we 
find A = | , B = — Y \, C = — T V, and k = T V . 



r dx 
J (14- sT' 



(1 + X*) 

du 



r,[K + **] + $ tan" 1 a; + a 



Here v = 1 4- a*, r = — 3, ^ = 2#, v = 1, and in order to 

dx 



INTEGRATION. 203 

express J ^ a . 3 in terms of J [= tan -1 x] we make 

v t = 1, s - — 1, s — r = 2, and (F) becomes 

1 = _ 4xf(x) + (1 + x*)f'(x) + &(1 + 2a; 2 + x*), 
where f(x) — Ax' + J9a;. 






In order to express the integral in terms of / 
[=log 



1- |/l - a; 2 % " 



make u~ r = xV 1 — a; 2 = yV — x\ 



— i, then u = x* — x\ r == — -J, -^ = 2a; — 4a; 3 , v - 



— — = 2x — 4a; , v — - 
aa; a; 

and (F) becomes 



ar 4 = («- 2a; 3 )/(a;) + (^ - a;<)/'(a;) + h, 
where f(x) = .^ar 6 -f Bx~ z -\- Cx~ l , since x* is a factor of u. 
f x 3 + 8a; + 21 , 

Here 

tt = z 2 _ 4a; + 9, r = - 2, ~ = 2a; - 4, and t; = »' + 8a; + 21; 

hence we have from (F), making s = — 1, 

a; 3 + 8a; + 21 = (4 - 2a;)/(a;) + (a; 2 - 4a; + 9)f'(x) 

+ ^(a; 2 - 4a; + 9). . (1) 

The second member of this equation must be of the third 
degree; but if we make v x = 1, the solution will be impossible, 
let us therefore assume that k = 1 and v 1 = Cx + D\ we may 
then make f(x) = Ax-\- B and f'{x) = A. Substituting in (1), 



204 DIFFERENTIAL AND INTEGRAL CALCULUS. 

we find ^4 = f , B — — *£-, 0=1, and D = y-. Hence the re- 
quired integral is 

2{x 2 -Ix-j-ty^J x* — 4x + 9 

3 (a; - 7) P (x - 2)dx P *£dz 

~ 2(x 2 — 4x + 9) + t/ a; 2 — 4a; -f 9 + t/ x* — 4z + 9 

The solutions of the three following examples illustrate the 
manner of integrating sin m x cos" x dx when the conditions 
stated in Art. 213 do not exist. 

5. / . . . -7-7 — ft cos 3 x — | cos afl 4- f log tan i x 4- C. 

J sm x sm 4 x L J ° 

Here we make m = sin x,r= — 5; then v=l and -=— = cos x. 

dx 

Making v 1 = 1 and 5 = — 1, (F) gives 

1 = — 4 cos xf(x) + sin xf'{x) -f- & sin 4 a:. 

Making sin 4 sc = (1 — cos 2 a;) 2 , /(a:) = ^4 cos 3 a; -|- J5 cos a;, 
and proceeding as in examples 6 and 7, Art. 213, we find A = %, 
B = — |, and & = f . 

.-. fj%- = -A-(i cos 3 s - I cos x) + f A^-. (Art. 204) 
«/ sm a; sm 4 ar y , «/ sin a; v ' 

This example may also be solved like the following one. 

e/ cos a; 

Make u = sin a?, r = 0, then ^— = cos a?, and v = cos" 7 x. 
.*. cos -7 x = cos a/(a;) -j- sin xf'{x) -f 7^, sin s a:, . (1) 



INTEGRATION. 205 

where f(x) = A cos -6 x + B cos * x + C cos -2 x, 

and /'(#) = ifiA cos -7 a; -f- 42? cos " 5 a; + 2 C cos -3 a;) sin a;. 

Substituting in (1), making sin 2 x = 1 — cos 2 a;, reducing, 
etc., we have 

6i-l,45-5i l 2(7-35 , " 

: — + 1 -4 § ■ 4- lev. sm s x — 0. 

cos a; cos a; cos a? cos x 1 

\ ^ = h B = fc <? = it> and (making 5=0, v 1 = ^- J * = a 

/^# sin a; . 5 . 1 sin x 

cos 7 a; ~~ 



cos 7 x 6 cos 8 x 6 . 4 cos 4 a; 

5.3.1 sin a: .5.3.1 



r_dx_ 
J cos a?* 



6.4.2 cos 2 a; 6 . 4 . 2 «/ cos a;* 
To integrate the last term see Art. 204, page 188. 

7. / sin 4 x cos 4 a; fe 

. „ f cos 6 a; , cos 3 a? 3 cos x~\ 3 , . „ 

gm .^___ + __ + _-J__( Bn *.oo B *-*)+a 

Make w = sin ar, r = 2, then v = sin 2 a; cos 4 x = cos 4 a;— cos 6 a:; 
also make 5 = 2 and v x = 1, then 

cos 4 x — cos 6 x = 3 cos a;/(a;) + sin xf'(x) -\-h. . . (1) 

Make /(a;) = ^4 cos 5 a? + J5 cos 3 x-\- cos ar, find f'(x), sub- 
stitute in (1), for sin 2 x write 1 — cos 2 x, etc., and we find 
A = -i, 5 = T V, tf= A, and £ = ^. 

/ 7 . a r cos 5 x , cos 3 a: 3 cos x~] 
sm 4 a; cos 4 xdx = sin a; — -\ — — | — — 

To integrate the last term, see ex. 11, Art. 185. 



206 DIFFERENTIAL AND INTEGRAL CALCULUS. 

x 2 dx 



8. 



Va? 



[- X 2 Va '-^ + Y^- lX aJ = ^ 



dx 



Va 2 - 



[ 



Va 2 - x 2 , 1 _ 



• a -L 



log (- 1) 



2a' 



10 



2a 2 x 2 2d 6 to a 4. |/ a * _ ^ 
. f Va 2 + a; 2 <fc. |Va« + af + y log (3 + W + z 2 ) + CI 

11. f(a 2 + x 2 )*dx. 

I (2a: 2 + 5a 2 )Vz 2 + a 2 + ^- log (3 + Vx T +~a~ 2 ) + a 

12. f\a 2 -x 2 fdx. %(5a*-2x*)VaT^* + ?fsm- 1 - + C. 
13. 



x*dx 



V2ax - x 2 



/»2a . ! 

14. / x'V 2ax — x 2 dx. 

n%a 

. / x % \2ax — x 2 dx. 
«/o 



= %na 



15 



16. 







\na\ 



x"dx 



Vi 



17 



18 



19 



2.4V2/ 

1.3.5 /tt\ 

2.4.6\2/' 



r x«dx 

' Jo ^F 7 ^' 

. / Vs + 2x -f x 2 dx. 
|/3 + 2x + ^i^^) + log (2a + 2 + 24/3 + 2a; + a; 2 ) + C. 

. y VlO + 3.r - x 2 dx. 

/ 1A , o i^a; - 3\ ' 49 . . 2z — 3 

|/10 + 3z - 2^— —J + — « 



-^ h 0. 



INTEGRATION. 207 



n 3x* - 2x + 5 



20. / dx. *£tt. 

oi C llx x l J_+ -i x n 

~ L J (« 2 + xY %a\a? + x*) + 2a 3 Un a + °' 

**' J (a; 2 + 3y 



* + ** 



— + o- log (a; 2 + 3) 



23 



4(z 2 + 3) 
' J (x* 



+ J-tan-^+C 
2 1/3 V3 



(a; 2 - 3a; + 3) 2 

1 3a: — 24 26 _ x 2a; — 3 

3p"-3a; + 3) + 3 ~^3 ~^~ + C * 

24. / sin 6 a^a;. 

sin 3 a; (J cos 3 a; — f cos x) — -f^ sin a; cos a; -j- -f^x + C. 

/sin 4 a; ., cos a; /sin 5 x sin 3 a; sin 2A , x „ 
— ~ dx ' — o— I" q 7^ q- + ?« + a 

sec a; 2 \ 3 12 8/16 

sin 4 a; sec -2 a: = sin 4 x cos 2 a; = sin 4 x — sin 6 a;. 

Hence we may make u = cos x, r — 0, and v = sin 4 x — sin 6 a\ 

/* sin 2 xdx sin a; sin a; . , , , , . , „ 

26. / 1 . i -5 5 i log (sec a; + tan x) -f (7. 

t/ cos a; 4 cos a; 8 cos a; ° v ' 

sin 2 a; _ 1 — cos 2 x _ 1 1 

cos 5 a; cos 5 a; cos 5 a' cos 3 a;' 

Hence we may make u = sin x, r = 0, and 2/ — cos -5 a; — cos -3 x, 

27. / cos 4 x cosec 3 a; fc — . . ' cos x — 4 log tan — 4- O, 

*J 2 sin's 2 & 2 

/* . e , sin 5 a; 5 sin 3 x , - r . n , ~ 

28. / sin x sec a; dx. 5 — — k§|a;— slna;cosa;H-C• 
«/ 3 cos 3 a; 3 cos a; 2L J 

29. / -^——dx. • — L . 3 V- 7 —-. sm x 

*J sm 4 x cos 2 aA3 sm x 3 sm x 2 

+ f log (sec x + tan a;) -J- C. 



208 DIFFERENTIAL AND INTEGRAL CALCULUS. 

215. Reduction Formulas for Binomial Differentials. 

These may be easily obtained by the method of indeterminate 
coefficients. 

I. Required / (a -f- bx n ) p x m dx in terms of 

J (a + dx n ) p x m - n dx. 

Make u = a + &% M > r =p, -f- = nbx 11 - 1 , v = x m , s — r, 
and v 1 = x m ~ n . 

Substituting in (F), we have 

x m = (p + Vjnbx'^fix) + (a + bx n )f'(x) + Jcx m ~ n , (1) 

where /(#) = Ax m - n+1 , and /'(#) = (m — w -f l)Ax m ~ n . 

Substituting these values in (1), arranging with reference to 
x, we find 

A = L and ]c = < m ~ n + x ) , 

5(wp + w + 1)' ' &(wp + Wi + 1)' 

Ci - ^ n\» «,j x m - n+1 (a + bx n y +1 
,\ / (a + bx n ) p x m dx = —n — \- ■ — — fr— 
e/ v ' ' b(np + m + 1) 

By a repetition of this formula m may be diminished by any 
integral multiple of n. 

II. Required J (a -\- bx n ) p x m dx in terms of 



/(■+ 



bx n ) p - l x m dx. 



Make w = a -f £# n , r =j? — 1, s =p — I, 

v = (a + bx n )x m , and v t = 2 W . 



INTEGRATION-. 209 

Substituting in (F) and proceeding as before, we get 

f(a + bx")Vdx = xm+1{a + *?}* 
J v J np + m + 1 

°L n P C(a _|_ lx«)*-H m dx. (B) 

Each application of this formula diminishes the exponent of 
a -f bx n by unity. 

When m or p is negative, we need formulas for increasing 
instead of diminishing them; hence the following : 

III. Required / (a -j- bx n ) p x m dx in terms of 

J (a + fo n ) V+ TC d£. 

Solving (A) for / (a + bx n ) p x m - n dx, and substituting ?ti -f- w 
for i», we get 

t/ v ' a(m -f 1) 

- »(«*> + » + "* + !) A,, + fc^-^fa (C) 

IV. Required I (a -\- bx n ) p x m dx in terms of 

J (a + fa n )* +1 a w rfa;. 

Solving (B) for J {a -j- &c B ) 3,-1 a: m (fc, and substituting i? + 1 
forjj, we find 

/ (a + bx n ) p x m dx = v ,\ ' — 

«/ 7 aw(p + 1) 

^ + ^+m+i /• j^^ (D) 



210 DIFFERENTIAL AND INTEGRAL CALCULUS. 

216. The approximate integral of many differentials may 
be conveniently obtained by the method of indeterminate coeffi- 
cients. The following important example will serve to illustrate 
the process. 

Integrate the Elliptic Differential 

dx 

Comparing this with (E), we may make u = a? — x*, whence 
r — — i, y- = — 2x, and v — (a 2 — eV)*, which, when devel- 

(XX 

oped by the Binomial Theorem, gives 

e* \ l_ \ e° , 
V ~ a 2a X 8a* X 16tf X ' ' ' 

Substituting in (F), Art. 211, making s = — -J, v 1 = 1, we 
have 

where f(x) is evidently of the form fx = . . . Ax b -f- Bx 3 -\- Cx. 
Proceeding as in Art. 212, we find 

•■•[•I- <••-»'[• ■«£)'+ (w+s)S" 

+ W + 64+4 AS +T-4" 64 ~ 256 " • "J Sm a" 



CHAPTER X. 

INTEGRATION AS A SUMMATION OF ELEMENTS. 

ELEMENTS OF FUNCTIONS. 

217. Hitherto nothing has been said about the magnitude 
of differentials. Whether they are large or small does not affect 
the principles which have been deduced; hence we may regard 
them as small as we please. They are variables whose limits are 
zero. 

218. In the present chapter increments are called and treated 
as Elements.* Thus Ay or Af(x) (— m J l + m Ji 2 , Art. 24) is an 
element of the function y — f(x). For convenience the element 
Af{x) will often be represented by E xy and the differential dy 
or f'(x)dx by D x , which may be called, respectively, the xth 
element and the xth. differential of the function f(x). Since 
D x varies as dx and approaches E x indefinitely as dx approaches 
0, D x is called the differential value of E x with respect to dx. 

The expression ^ 2 [E X ~\ represents the sum of all the ele- 

ments like E x , or the sum of the successive values of E x , be- 
tween the z-limits a?, and x r That is, supposing the increment 
of x to be always li, 

2 x [Ex\ = E Xl -\- E Xl + h + E Xl + 2h + • • • E x ^ _ h (or Xl + ( n _ 1)70. 

219. A Definite Integral Regarded as a Sum. The prac- 
tical importance of integration consists chiefly in regarding it 

* Because sum and element are correlative terras. 

211 



212 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



as the summation of a certain series. For example, in seeking 
the area of a curve, we conceive it divided into an indefinite 
number of suitable elementary areas, of which we seek to deter- 
mine the sum by a process of integration. The solution of this 
fundamental problem is effected by the following formula and 
its corollaries. 

Suppose that in any function of x, as/(#), we change x from 
x x to # 3 by giving to x successive increments. The whole change 
in the value of f{x), viz.,/(# 2 ) — /(#,), must be the sum of the 
partial changes produced by the increments given to x. That 
is, 

or (Art. 208) f**D x = ^\E X ] (A) 

Formula (A) is not only an expression of the simple fact 
that the whole is equal to the sum of its parts or elements, but 
it signifies that the integral of the differential value of an 
element, between certain limits, is the sum of the successive 
values of that element, between the same limits. 




Fig. 43. 

As an illustration of (A) let us consider the area of BGQF y 
where OE — x 1 and OG = x r Suppose y=f'(x) to be the 
equation of AFQ, where x = OB and y = BP. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 213 

Let u = the area of OB PA, and let BC ( = h) be an element 
of x, then BCDP - du =f'(x)dx = D x , and BCP'P = D x + 
au = &. 

(a) Evidently, EGQF = f X *B x , Art. 208. . . . (1) 

(#) Divide i?6r (= x 2 — aj into w parts, each equal to li, and 
draw the ordinates a 1 d 1 , a 9 d 2 , a 3 d 3J etc., then 

EGQF=Ed 1 + a 1 d u + a,d 3 +, .... (2) 

in which Ed l , a x d 9 , # 2 d 3 , etc., are the successive values of E Xi as 
a increases by h from a^ to x 9 . That is, Ed x —E Xx , afi^—E^ + hy 
a 2 d 3 = E Xl +2h> etc. Hence (2) may be written 

EGQF=^[E X -] (3) 

Now, equating (1) and (3), and we obtain formula (A). 

In further illustration of formula (A), let us show that the 

signification which it expresses is true of / 3x*dx. 

tJX\ 

(a) r x *$x t dx = \x l + Cf* = xf-x l \ ... (4) 

t/Xi X\ 

(b) Since /'(a) = 3# 2 , E x = 3x*h + Zxh* + h% Art. 207, 

E Xi =3x*h + 3x 1 h 2 + h 3 ; 
E Xl + h = 3x x *h + 9xjf -f W; 
E Xl + 2h = Sxfh + lbxji* + 19& 3 ; 



E Xl + (n -i)fc= 3?^ + 3^(2^ - 1)A" + (3^ 2 -- 3» + 1)£ 3 . 

Taking the sum, remembering that x x + nil = x 9 , and, by 
Algebra, 3 + 9 + 15 + . . . 3(2rc - 1) = 3rc 2 , and 1 + 7 + 19 + 
. . . (3w a — 3n -f 1) — n a , we have 



^[^J = 3nhx* + 30^)^, + (wA) 1 



V 

= (a;, + fiA) 1 - a:, 3 = z 2 3 - x*. ... (5) 



214 DIFFERENTIAL AND INTEGRAL CALCULUS 

Comparing (4) and (5), we see that the results of the opera- 
tions indicated in (A), when applied to / Sx^dx, are the same. 

220. Formula (A) is also true when x 2 — x x (— EG) is not 
divided into equal parts. 

Let us suppose x 2 — x x to be divided into the following equal 
or unequal positive parts : 

a — x lf b — a, c — b,.. . . I — h, x 2 — I, 

the sum of which is evidently x 2 — x } ; then we have identically 

/ y dx — I y dx -f- if dx-\- y dx -\- . . jydx > (1) 

t/ari vxi *Ja *" Jb Ok 

in which a — x l9 b — a, c — b, etc., may be considered the suc- 
cessive values of Ax and / y dx, I ydx, etc., the correspond- 
ing successive values of E x . Hence (1) is the general significa- 
tion of (A), which the student may easily illustrate with a 
figure. 

221. A Definite Integral Regarded as the Limit of a Sum. 

In Fig. 43 let us suppose n to increase and h to decrease, nil 
being always equal to EG. Since the limit of E x -±- D x , as h 
approaches 0, is unity, the sum of all the rectangles like D x 
approaches indefinitely the constant sum of all the elements 
like E x . Therefore 

limit ^*F-n 1 ^Trrl 
Substituting in (A), Art. 219, we have 

i>="S<M <»> 

That is, the definite integral / D x is equal to the limit of 

the sum of all the successive values of D x , as x increases by h 
from x, to x„. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 215 

For example, let us find the value of / 3x 2 dx. 
Since D x = Sx^h, we have 
D Xl = Zx?h, 
D Xl + h = Zv*h + 6*^" + U\ 
D Xi + 2h = 3x^h + 12xJ? + 12h% 



D Xl + {n _ 1)h = Zx?h + Q(n - l)xfi' + 3(w - 1)'*\ 
Taking the sum, remembering that 

+ 6 + 12 + . . . 6(n - 1) = 3(?z 2 - ri), 

and + 3 + 12 +» + . ■ . 3(n - 1)« = (W " 1} ^ " X \ 
we have 

^>,] = Sate.! + 8(»- - »)*,*• + (n ~ 1){n) 2 (2n ~ 1] h> 

= ${iih)x? + 3(^)X - 3(wA)^ + {nhy - \{nlifh +- (^V. 

Now, making nh = x^ — x 1 , and then passing to the limit 
by making h = 0, we have 

j^ 2 x 'P*l = 3 ^ - *>.' + 3 ^ - *.)"*. + fo - *,) 3 

which is evidently equal to / Safdx. 

222. It is important to observe that, whether an integral be 
regarded as a sum, or the limit of a sum, integrating is equiv- 
alent to two distinct operations: 

(a) If a st«m, as in Art. 219, integrating/' (a:) $» is equivalent 
to (1) increasing the differential f'(x)dx by the acceleration au 
to obtain the element E x , and (2) finding the sum of the succes- 
sive values of E~ 



216 DIFFERENTIAL AND INTEGRAL CALCULUS. 

(b) If the limit of a sum, as in Art. 221, integrating/ '{x)dx 
is equivalent to (1) finding the sum of the successive values of 
f'{x)h, and (2) taking the limit of the sum, as h approaches 0. 

In case (a) all the quantities involved are finite; but in case 
(5) the limit of each part is 0, and the limit of the number of 
parts is qo . Both methods have their advantages, and hence both 
will be employed, more or less, in the applications which follow. 

Just here the student may profitably read Art. 238, which 
offers a simple illustration of the significations and practical im- 
portance of formulas (A) and (B). 

APPLICATIONS TO GEOMETRY.* 

223. Lengths of Curves.— I. Bectangular Co-ordinates. 

To find the length (s) of the arc APQ between the limits 
OB = x x and OG = x v (Fig. 44.) 

d 5 




Fig. 44. 



Here E x = Pd x and D x = Ft = /l -f (^f) V Art. 33, and 



* The previous applications of Calculus to Geometry, Arts. 62, 64, 65, 
66, were limited to the most elementary rules for integration ; in this 
chapter it is our purpose to extend these applications by the more advanced 
methods of integration with which the student is now familiar, and in doing 
so to impress upou him the important principle of integration as a summa- 
tion. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 217 
#2 limit - r 2 P%i 

Pd, + d,d, + a A + etc. = 2 Xi W\ or ^^[2).]=^ Z>*. 

- -XViTS)^ or £V^(§)W. ( C ) 

EXAMPLES. 

£ 3 1 

1. Find the length of the arc of the curve y = — + — be- 
tween the limits x, = 1 and x a = 2. 

dx 



dy _ x* - 1 / <fy'\*__ 1 +^ 4 _ rfo 
Mere tfz~ 2z 2 ' V + drf)~ 2x> "da 



2*? + ^ "2- 



~ L 2» 4 



~6Ji 



2. Rectify the parabola y* = Aax, using the formula 
dy y \dyH 



dy~2a' •■dy- y 4a' + 1 -2a iy +ia - 

= y V4g + 7 +gb8 / |>+Vg +g\ > Art. 214, Ex. 10. 

3. Bectify the curve y = log (x + |/x 2 — 1). 

4. Rectify the ellipse ?y 2 = (1 - e 2 )(a 2 - x 2 ). 



dy ,„ ..a; xVl-e 2 

-j^- ■= — (1 — e 1 )- = 

<fc «/ tV - # 2 



218 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



To find the length of a quadrant, we must integrate between 
the limits and a; hence 



-/v. 



x 2 (l -e 2 ) 
a 2 - x 2 



dx 



= £ ♦*»■ - *'■■ 



dx 



- Eh _ 1 • _ 1 - 3 * _ 1 - 32 • 5 

~ a 2\ 4 6 2 2 .4 26 2 2 .4 2 .6 2 



Va*- 
— etc. J. 



x 

Art. 216. 



5. Kectify the cycloid x — r vers -1 - — V 2ry — y*. 



Here 



dx 



y dy 



V2ry-f 9 



yW 



V- 



2r 



ds = \Uf , 

^ 2r?/ — if J T 2r — y 

s = J (2r)\2r -t J y i dy=-2 |/2r(2r - y) + 01 

If we estimate the arc from the 
point i? where ?/ = 2r, we shall 
have, when s = 0, y = 2r; hence, 

= + C, .\ C=0, and 




Since BO = 2r and BE 



= - 2 V2r(2r - y). 
t, BG= V2r(2r - y). 



BD=s = 2BG; 

or the arc of a cycloid, estimated from the vertex of the axis, is 
equal to twice the corresponding chord of the generating circle; 
hence the entire arc BDA is equal to twice the diameter BO, and 
the entire curve ADBH is equal to four times the diameter of 
the generating circle. 



6. Rectify y = -^ + ^ 



16 



2z 2 



+ 0. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 219 

Kote. — The value of C depends on the point from which s is 
measured. Thus, if s is estimated from x = 1, then 5 = when 
x — \, and we have = ^ g — \ -f C; that is, C = T \. 

x n+1 1 

7. Rectify y = — — - + 



4(^ + 1) ' (w-l)^"- 1 



*^-^-V- i+a 



v 2-m 



8. Eectify y = — -. s = y + + 0. 

J * 2w 2(2 — m) J ' 2 — m 

9. Eectify y = -J log (z 2 + 3a: + 2). 

10. Rectify ?/ = J# 2 — log x. s = \x 2 -\- logx -\- C. 

A curve is said to be rectifiable when its length can be ex- 
pressed in finite terms by aid of the algebraic and elementary 
transcendental functions. 

224. II. Polar Co-ordinates. To find the length (s) of 
the arc APQ between the limits 6 1 = AOP or r 1 — OP, and 
2 = AOQov r^= OQ. (Fig. 46.) 

Here E» = Pd x and D 9 = Pi l = \^ r *+^ de > Art - 97 '> 
hence 

Pd, + d x d % + d 2 d 3 + etc. = 2*[Ee] 

limit A n1 /"'mi 

- -=r(-+s)>-/X i+ ^') v - (D) 

11. Rectify the spiral of Archimedes, r — aO. 

Heref = 1; ,. s = /*(l + 3 W * /> + O** 

dr a e/o V ' «v « t/o v ' y 

r(« 2 + r 2 )* , a . r + ^<?+7* A x 01 , -c ^ 
= ^ + j log ^ g T„ Art. 214, Ex. 10. 



220 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Let this result be compared with the length of the parabola, 
Ex. 2. 

12. Find the length of the logarithmic spiral log a r = 0, 



Here 
and 



dO 



mclr 



eld m 



r 



r ' dr 

s = f Q (1 + m'fdr = r(l + m*)* 




13. Find the length, measured from the origin, of the curve 



y = a log 



a* — x' 



. a + x 
a log - 



a? " a — x 

14. Find the entire length of the arc of the hypocycloid 

x % -f y l = a\ 6a. 

15. Find the entire length of the cardioid r — a(l — cos 6). 

8a. 

, - , . , # 37r« 

16. Find the entire length of the curve r — a sin -. — ^-. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 221 

17. Find the "length, between x — a and x = b, of the curve 

e x + 1 , e 2b - 1 , 

18. Find the length of the tractrix, measured from (0, a), its 
differential equation being 

ds a . a 

-j- = . c log -. 

dy y *y 

19. Find the length of the arc, measured from the vertex, of 
the catenary 

c( T - - x \ c( x - - x \ 

y = 2\ eC + e c )> 2\ eC ~ e e h 

20. Find the length of a quadrant of the curve 

/*y , toy. a a +«ft + y 

[a J ^Kb) - U . « + 6 * 

225. To find the equation of a curve when its length is 
given. 

1. Find the equation of a curve whose length is 

x 2 
s = |logz+-. 



ds 
dx 



1 + ^. . i/(d s Y l — x* _dy 

2x ~ ; •"' * \dx) ~ ~ 2x ~~dx' 



1 — x* 
Hence, dy = — - — dx, and y = i log z — \x* -f- C 



EXAMPLES. 
Find the equations of the curves whose lengths are: 

*•* = -■& + £• y = s ~^ + C - See Note, p. 219. 



222 DIFFERENTIAL AND INTEGRAL CALCULUS. 

'x-Y 



3;.* = *+-l0 gVr + ]y 

4. s = i log (tan #) 
1 



y — \ log (sin 2#) -j- C. 



5. 5 



i + 



2(n- l)^- 1 * 2(» + l) 



y = s- 



n-\-\ 



+ 0. 



226. Areas of Curves.— I. Rectangular Co-ordinates. To 

find the area of the surface between a given curve, the axis of x 
and two ordinates whose abscissas are x l and x t , we have, Art. 
219, 

«■=./ ydx = 2 [E x ]. . . ; . . (E) 
For a definite area between the curve and axis of y, we have 
u= xdy = 2JEy\ (F) ' 

EXAMPLES. 

x* 1 
Find the area of the curve y = —— -j- .—— between the limits 

x t = l and # 2 = 2. 

I " f x* i \ r~# 6 i n 2 51 

fo = i lie- + §#r* = Lio- - & J, = 80- 

2. Find the area of the circle y 2 = a 2 — x 2 . 

Area of OB PA = fydx 

= / (a 2 — x 2 Ydx = — — - — *- -f — sin -1 - 
t/o 2 2 rt 

(Ex. 17, Art. 96) 




X B C 

Fig. 47. 



_ (0B)(BP) (0P)(slycAP) _ 



+ 



= area 0£P 4- area OP A. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 223 
To find the area of the quadrant OCA we have 

f\a 2 - tffdx - ±na\ 

I/O 

The value of n is given in Art. 129. 

3. Find the area of an ellipse, a 2 y 2 = cTb 2 — b 2 x 2 . 

y = - l/V - x 2 ; .'. u=- fv'a^^x 2 dx. 

i (the area) = -*- / Va 2 — x 2 dx = \nab\ 

entire area = nab. 

4. Find the area of the hyperbola, a*y* = b 2 x 2 — a 2 b 2 . 



y 



= - Vx 2 — a 2 : ,\ u = - / \/x 2 — a 2 dx. 
a a J 



or 



bx{x 2 - a 2 ) k ab . , , A/ - 2 5 x , n 

u = -A_^_ — log (x + Vx 2 - a 2 ) + 0. 

To find C, we know that when x — a, u = 0; hence 
0=- a poga+C; .: C = ~loga. 

Substituting this value of C, and making Vx 2 — a 2 — -~- f 



we have 



Area of hyperbola = -^ - ~ log (^ + 1 J. 

5. Find the area of the surface between the arc of the pa- 
rabola y 2 = ±.ax and the axis of y, \xy. 



224 DIFFERENTIAL AND INTEGRAL CALCULUS. 



2*21. It is often convenient and suggestive to regard a defi- 
nite integral like 

f Z 'y(dx) = EGQF (Fig. 43) 

as signifying that "the ordinate y 
(or generatrix PB), moving perpen- 
dicularly to the axis of x from x=x x 
to x = x 9 , generates EGQF. 

Thus, let it be required to find 
the area of the surface between the 
parabola y 2 = 4ax and the straight 




Fig. 48. 
line y = x. We at once have 



OP DC 



= / (PC)dx = / (V 

*J 1/0 



4ax — x)dx 






This method can be employed, with equal facility, in finding 
the volumes of many solids, in which case the generatrix is a 
surface. 

6. Find the area of one branch of the cycloid. 



dx 



_yfy . . u= _ r tfdy 

V2ry - f~' J V2ry-y* 



For the area of one branch we have 

= %7ir\ (Ex. 13, Art. 214) 



/*Zr 

J n V2rv- 



-y-y 

which is three times that of the generating circle. 



'. Find the area of the curve xy 



a. 



M 



a log 



8. Find the area of the curve x 5 y — x + 1 = between the 
.^-limits 1 and 2. £. 

9. Find the area of both loops of a*y* — aWx* — b*x\ %ab. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 22b 
10. Find the area of both loops of the curve a e y 2 = a V — x*. 



11. Prove that the area of the curve a^y 3 = (a 3 — x 3 )x\ be- 
tween the ^-limits and a, is the same for all values of a. 

228. II. Polar Co-ordinates. In Fig. 46 the area of POD 
(— ±r*dd, Art. 35) is the differential value of the element POd/, 
therefore 

jf V<W = P0d x + dfid, + dftd % + . . . dUflft 
where 6 = ^4 OP and 2 = ^0§. 

u = ij e *r*d6 = area POQ. (G) 

12. Find the area of the spiral of Archimedes, r — ad. 

du = — ; .-. u = rr— I r dr = -— . 
a ' 2a t/o Qa 



Cor. I. If a — — , as is usual, u — \nr 3 . 

If r = 1, or 6 = 2n f u = -|7r, which is the area described by 
one revolution of the radius vector. 

If r = 2, or 6 = In, u = %n, which is the area described by 
two revolutions of the radius vector, which includes the first 
spire twice; hence the area of the entire spire is %n — \n = in. 

13. Find the area of the hyperbolic spiral, r = — . 



au 






14. Find the area of the logarithmic spiral = \og a r. 

d6 = ; .*. when m = 1, 

r 

du = —jr- and u = ^r 3 ; 



226 DIFFERENTIAL AND INTEGRAL CALCULUS. 

that is, the area of the natural logarithmic spiral is equal to one 
fourth the square described on the radius vector. 
15. Find the entire area within the hypocycloid 

x + y* — a • f na % 

1.6. Find the area of the surface between the parabola 
x* = 4:ay and the witch 

ft/7 3 

17. Find the entire area of the cardioid 

r = a{\ — cos 6). f 7ra\ 

18. Find the area of a loop of the curve 

x* -\- y* = a a xy. ^na*. 

19. Find the area of the loop of the curve 

a y = x\b + x). 



326 



105rt § 

20. Find the area included between the axes and the curve 

\a) ^Kb) ~ ' 20* 

21. Find the area between the curve x 7 y* + #V = a *% 7 and 
one of its asymptotes. 2a*. 

22. Find the area of the loop of the curve 

y 3 + ax 2 - axy = 0. £ 6 a\ 

23. Find the area of the three loops of the curve 

r = a sin 30. (See Fig. 38.) ina\ 

24. Show that the whole area of r = a(sin 2d -f- cos 26) is 
equal to that of a circle whose radius is a. 

229. Areas of Surfaces of Revolution. By Art. 34 the 
differential value of the xth element of surface is 

s *y( i +.K**»i 



INTEGRATION AS A SUMMATION OF ELEMENTS. 227 
therefore 

r x2 

S = 2*J^ y(l -f g)\fe = 2j[jy- • • (H) 

EXAMPLES. 
1. Find the area of the surface generated by revolving the 

x* 1 
arc of the curve y = — + =-= about the axis of #, between 
■* 16 2x 2 

x i = 2 and x 2 = 4. 



i/ 1 + (^y = ^ = ^! + i 

r T Wz / <fc 4 T z 3 



4 «-S+ &)(?+> 



[•te+S-y+'^lr" 7 *^- 



2. Find the area of the surface of a prolate spheroid, the 
generating curve being the ellipse a 2 y 2 = b 2 (a 2 — y 2 ). 



y = — Va 2 — x*. and ds = \ — 5 ~dx. 

9 a r a 2 — x 2 

.'. Area = 2 / 2;ry^s = 47T- / (a 2 — e'x*)*dx 
Jo a Jo 

. b [~ x(a* - e'x 2 )* , a 2 . . ex~] a 
— 4^_ _v /, ^_ sm -i _ 

a L 2 2e a J 

= 2tt6 2 H sm _1 e. 

e 

3. Find the area of the surface generated by the revolution 
of the cycloid about its base. 



/*2r /*2r 

Area = 21 2nyds = 47T 4/2 r / 
*/o Jo 



o V2r -y 



228 DIFFERENTIAL AND INTEGRAL CALCULUS. 



it y2r£- f(4r + y)(2r - y?~j 



64 ■ 
= T /Tr ' 

4. Find the surface of the paraboloid between the limits 
x = and z = a, the generating curve being ^ 2 = 4o& 

i(V8- l)87ta\ 

5. Find the surface generated when the cycloid revolves 
about the tangent at its vertex. - 3 ¥ 2 -arr a . 

6. Find the surface generated by the revolution about the 
axis of x of the portion of the curve y = e x , which is on the left 
of the axis of y. n {y% _j_ i og (i + 4/2)). 

7. Find the entire surface of the oblate spheroid produced 

by the revolution of the ellipse a?\f -(- #V = a 2 b 2 about its minor 

axis. ' b 2 1 + e 

2^-rt -f n— log; — - 1 — . 
1 e 5 1 - e 

8. Find the surface generated when the cycloid revolves about 
its axis. 87rr 2 (7r — J). 

9. Find the surface generated by revolving the arc of the 
cardioid r = a(l — cos 6) about the initial axis. 

8 = Infyds = 2?tfr sin 8 Vdr* + rW. ¥*«*• 

10. Find the surface generated by the revolution of a loop of 
the lemniscate r 2 = a 2 sin 26 about the polar axis. 2t« 2 . 

230. Volumes of Solids of Revolution. In Art. 32 the 
volume of the cylinder {ny^dx) generated by the revolution of 
the rectangle BCDP is the differential value of the xt\\ element 
of volume; therefore 

v = 2%[M«] or J™" 2 ?**"*] = *£?*' d *- (J) 



INTEGRATION AS A SUMMATION OF ELEMENTS. 229 

EXAMPLES. 

1. Find the volume of a paraboloid, the generating curve be- 
ing the parabola y 2 = kax. 

v = n j y 2 dx = 4:7ta I xdx = 2nax 2 . 

When the curve is revolved about the axis of y, we evidently 
have 

v — n j x 2 dy. 

2. Find the volume generated by revolving the surface be- 
tween the parabola y = + Vkax and the axis of y about that axis. 

[vf o = nfx'dy = nf^dy = £jL = ^x'y. 

That is, the entire volume is one fifth of the volume of the 
circumscribing cylinder; therefore the volume generated by the 
surface of the parabola in revolving it about the axis of y is four 
fifths of that of the circumscribing cylinder. 

3. Find the volume of the solid generated by the revolution 
of the cycloid about its base. 

dx = — -/ • ; .*. dv = ny 2 dx = 



V2ry — y 2 V2ry — y 2 

To obtain half of the volume, we must integrate between the 
limits y = and y = 2r. 

Jo V2ri J - f 
That is, v = f of the circumscribed cylinder. 
4. Find the volume generated by revolving the ellipse AA f 
about the tangent X'Xas an axis. (Fig. 49.) 

Let OA=a, OY=l, O'B = x, BP = y, and BP' = if \ 



than y = -(a -j- \/a 2 — x 2 ), and y' — -(a — Vet 2 — x 2 ),. 



230 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



Supposing P'n = dx, the volume generated by P'm, viz., 



n(y i — y n )dx or 47r- | 7 a 2 — x*dx> 



-x- 



y^ 


X 


m 


/ 






A 


\ ° 






J 


x^ 


p' 


y 


4 

n 


^^^^__ 





O' 

Fig. 49. 



is the differential value of the volume of the element generated 
by P'cdP, with respect to dx. 



f 



±n- Va? - xhlx = 2ttW, (Ex. 17, p. 56) 



which is the entire volume, being the sum of the volumes gen- 
erated by revolving all the elements like P'cdP between x— —a 
and x — a, or the limit of the sum of the volumes generated by 
all the rectangles like P'nmP. 

5. Find the volume of the closed portion of the solid gener- 
ated by the revolution of the curve («/ a — by = « 3 ^ around the 
axis of y. 256 nV 

315 « 6 * 

6. Find the volume generated by revolving the curve 
(x — 4a)y 2 — ax(x — 3a) about the axis of x, between the 



^-limits and 3a. 



^-(15 -16 log 2). 



7. Find the volume generated by revolving the cycloid round 
the tangent at the vertex. n' 2 r 3 . 

8. Find the volume and surface of the torus generated by 
revolving the circle x' 2 -f- (y — bf = a 2 about the axis of x. 

Zn-cvb and An^ab. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 231 

9. Find the entire volume and surface generated by revolv- 
ing the hypocycloid x* + y* = c? about the axis of X. 

32tt0 3 _ 127m 2 
and 



105 5 

10. Find the volume generated by the curve xy* = 4a 2 (2a— x) 

revolving about its asymptote. <±/r 2 # 3 . 

x 

11. One branch of the sinusoid ?/ = I sin — is revolved about 

the axis of x; find the volume generated. %7r-ab'\ 



SUCCESSIVE INTEGRATION. 

231. A Double Integral is the indicated result of reversing 

cVu 



the operations represented by 7 , » 

Thus, if -j— -y- — xy*. then u — I I xy"dy dx, 

which indicates two successive integrations, the first with refer- 
ence to x, regarding y as constant, and the second with reference 
to y, regarding x as constant. 

Thus, ffxtfdy ** =f[j + c )y* d y 

where C and C x are the constants of integration. 

232. Definite Double Integrals. Here both the integra- 
tions are between given limits. 



Pc Pa 

For example, / J x*y*dx dy. 



This notation indicates that the integrations are to be taken 
in the following order : 



232 DIFFERENTIAL AND INTEGRAL CALCULUS. 

J, Jo *Ydxdy = J b [J o xYdy)dx 

=i'(¥)*=^'-»' >• 

That is, as dy is written last, the ^-integration is taken first. 
The limits of the first integration are often functions of the 
second variable. 
For example, 

/a /» Vx n a 

I dx dy- I ( Vx )dx = fa 1 . 
Jo Jo 

As another example, 

rdrd9 = J Qdr = A b\ 

233. A Triple Integral is the indicated operations of three 
successive integrations, for which the notation is similar to that 
of double integrals. Thus, 

£ a £fJ x y zdxd y ih =X\Jo\fJ x y zdz ) d !i'] dx - 
= £ a \J, X{ix * yi)dy ~] dx = J? ww* = ^ a °- 

EXAMPLES. 
Find the following : 

L jf jf>° + y,)dx d,j - ir,b{a ' + 6> >- 




~- / / {f+f)lxdy. \na\ 

Jo Jo 



3. 


2a l 


'a 

I 


rVa?-ax 

1 clx clz 






^ Vax - 


-X* 

x* 4- ?/ 2 
a 

dxdy 




4. 


•VOax-x* / 


clz. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 233 

lna\ 

AREAS OF SURFACES DETERMINED BY DOUBLE INTEGRATION. 
234. Plane Surfaces — (a) Rectangular Co-ordinates. In 
the formula u = I y dx, Art. 226, we may make y = I dy, 
which gives 

u = I j dxdy (J) 

/r 2 





F 


> '—"- — >»^^ 


v/^ 


D >v 


/ q 




P \ 








n \ 


/ m 





B C 



Fig. 50. 



Art. 228, we may make - = / rdr, and write 
u = I I rdddr. . . 



• • (K) 



234 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



As illustrative examples let us employ (J) and (K) in finding 
the area of a circle whose radius is a. 

Let (x, y) be the co-ordinates of the point m, and (x -f- dx, 
y -\- dy) of the point p, then mnpq = dxdy. 

Kegarding x as constant, we have 

v=bp 
BGDP = 2 [mnpq] 
y=o 

\/2ax-x* 



/» \2ax-x* 

= dx I dy = \2ax — x 2 dx. 



Again, since V 2ax — x*dx is the differential value of BP' 
with respect to dx, we have 



2a p- a _ 

? [BP'] = / V2 

^0 



ax 



x 2 dx = in a? = area of OKP. 



(b) Let (r, 0) be the co-ordinates of m, and (r -\- dr, -\- dd) 
of //, then mn = rfr, wg = w?0, and rdddr ( = area of mnpq) is 

J 3 / 




Fig. 51. 
the differential value of the element mnp'q' with respect to dr 
Therefore, regarding 6 as constant, we have 

2 [mnp'q'] = I rdddr = 2a 2 cos 2 Odd = 0PP x '. 
o * 



INTEGRATION AS A SUMMATION OF ELEMENTS. 235 

Again, since the area of OPP/ is the differential value of 
the element OPP', with respect to d6, we have 



2*[0PP'] = f 2a" cos* Odd = ina\ 
which is one half the area of the circle. 



EXAMPLES. 

1. Find the area (1) of a rectangle by double integration; 
(2) of a triangle. 

2. Find the area between the parabola «/ 2 = ax and the circle 
y* = 2ax — x\ J7ta* 2a 2 \ 

3. Find by double integration the entire area of the cardioid 
r = a(l — cos 0). Zna 2 

~2~' 

4. In a similar manner find the entire area of the Lemniscate 
r* = a 2 cos2d. a\ 

5. Find the whole area between the curve xy' 1 = 4a 2 (2a — x) 
and its asymptote. ^na?. 

236. Surfaces in General.— To find the area (= S) of a 
surface whose equation is f(x, y y z) = 0. 

Let (x, y, z) be the co-ordinates of any point P of the surface, 
and (x -f- dx, y -f- dy, z + dz) the co-ordinates of a second point 
Q very near the first (Fig. 52). Draw planes through P and Q 
parallel to the planes XZ and YZ. These planes will intercept 
a curved quadrilateral PQ on the surface; its projection pq, a 
rectangle, on the plane of XY] and a parallelogram p'q' , not 
shown in the figure, on the tangent plane at P, of which pq 
is the projection. The area of p'q' — dS, since it is the dif- 
ferential value of PQ (= AS) with respect to dx and dy. 

The projection of p'q' on XY is dxdyx similarly the projec- 
tions of p'q' on XZ and YZ are dxdz and dydz; hence, denoting 



236 DIFFERENTIAL AND INTEGRAL CALCULUS. 

the angles between the plane otp'q' and XY, XZ and YZ by 
a, fi and y, respectively, we have 

cos adS = dxdy, (1) 

cos fidS = dxdz, (2) 

cos ydS = dydz (3) 




Fig. 52. 



Squaring (1), (2), (3), and adding, remembering that 
cos 2 a -f cos 2 /3 -f cos 2 y = 1, 



we have 



hence, 



(dS)' = dx'dy* + dx*dz> + difdz*; 






INTEGRATION AS A SUMMATION OF ELEMENTS. 237 

EXAMPLES. 

1. Find the area of one eighth of the surface of the sphere 
x- + y 2 + 2 a = a\ 

„ dz x dz v 

Here -j- = , -7- = - -. 

dx z dy z 

"^ £&* "^ <ty a ^ Z 2 "^ 2 2 2 a ' 

Substituting in (L), we have 

~_ P Padxdy _ P P dxdy 
~ J J z a J J */ a * - x 2 - y*' 

Integrating first with reference to y between the limits 
y = and y = Va 2 — x", we get the differential value of the 
element B'C'KL\ and then integrating with reference to x 
between the limits x = and x = a, we get the sum of all the 
elements like B'C'KL between these limits, which sum is the 
area required. 

Hence S=a / dxdy = 7ra> 

Jo Jo Va 2 - x 2 - y 2 ' 2 " 

2. The two cylinders x 2 -f z 2 — a 2 and x 2 -\- y 2 = a 2 intersect 
at right angles; find the surface of the one intercepted by the 
other. 8a 2 . 



Here z = Va 2 — x 2 . and for one eighth of the required sur- 
face the ^/-limits are and Va 2 — x 2 , and the ^-limits and a. 
3. A sphere whose radius is a is cut by a right circular 

cylinder, the radius of whose base is — , and one of whose edges 

passes through the centre of the sphere; find the area of the 

surface of the sphere intercepted by the cylinder. 2a 2 (n — 2). 

Take x 2 + y 2 -\- z 2 = a 2 for the sphere, and x 2 -f y* = ax for 

the cylinder, then z = Va 2 — y 2 — x 2 , and for one fourth of the 



238 



DIFFERENTIAL AND INTEGRAL CALCULUS. 



required surface the limits of y and x are 0, Vax — x 2 , and 0, a, 
respectively. 

4. In the preceding example find the surface of the cylinder 
intercepted by the sphere. (See Ex. 3, Art. 233.) 4a 2 . 

5. Find the area of the portion of the surface of the sphere 
x % -j- y" 2 -f- z 2 = 2ay lying within the paraboloid y = mx 2 + nz 2 > 

27ta 
Vmn 
237. To find the volume of a solid bounded by a surface 
whose equation is f(x, y, z) = 0. 




Fig. 53. 

VOLUMES OF SOLIDS DETERMINED BY TRIPLE INTEGRATION. 

Let v = the indefinite volume expressed by the product of 
x, y, z; then v = xyz, which may be written 



v = I I I dxdy dz, 



which becomes definite when the integrations are taken between 
certain limits, and we will now give the geometrical interpreta- 
tion of the formula, step by step. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 239 

Let (x, y, z) be the co-ordinates of the point P, and (x -f- dx, 
y -f dy, z -f- dz) be the co-ordinates of the point Q; then 

PQ = dxdy dz. 

(a) Kegarding x and y as constant and integrating between 
the 2-limits and id, we have 

2 M [PQ] = J^dxdydz = (id)dxdy=M. 

(b) The volume of bh is the differential value of the element 
is' with respect to dy, hence 

nam 

2™ [is'] = J o (id)dx dy = (afm)dx, 

which is the volume of the cylindrical segment afm-a'. 

(c) The volume of afm-a' is the differential value of the 
element ar with respect to dx; hence 

^ o \ar] = I (afm)dx = volume of OBC-A. 

pOA nam r*i& 

J o J o J o dxdydz — volume of OBC-A. . . (M) 

Cor. I. The limits of y and z are found thus: id is the posi- 
tive result of solving the equation f(x, y, z) = for z, am is the 
positive result of solving f(x, y,0) = for y, and OA is the 
positive result of solving f(x, 0, 0) = for x. 

EXAMPLES. 
1. Find the volume of one eighth of the ellipsoid 

a* ^ b* ^ c* 

I x* ?/ 2 \* 
The limits of z in this case are and id = ell i* — fr) 5 



240 DIFFERENTIAL AND INTEGRAL CALCULUS. 

the limits of y are and am = bi 1 -j ; and the limits of x 

are and a. Therefore the required volume is 



J. 



V a2 r c * l a i ; j2 

/ dx dy dz = 



6 

2. Find the volume of the solid contained between 

(a) the paraboloid of revolution, x* -\- y 2 = az, 

(b) the cylinder, x 1 -f y' = 2ax, 

(c) and the plane, 2 = 0. 



37ra a 



2 " 

# 2 4- ?/ 2 
The 2-limits are and — J , the y-limits are and 

V2ax — x 1 , and the ^-limits are and 2a, for one half of the 
required volume. 

3. Find the volume cut from a sphere whose radius is a by a 
right circular cylinder whose radius is b, and whose axis passes 
through the centre of the sphere. irf . 3X 

s F x(« ! - («' - *)')• 

4. Find the entire volume bounded by the surface whose 

equation is x* -f- y* + s? = a 1 . 4?rfl 3 

35 ' 

5. Find the volume of the conoid bounded by the surface 
z*x* + a*y* = cV, and the planes x = and x = a. %itac\ 

APPLICATION TO MECHANICS. 
238. "Work is said to be done when a body moves through 
space in opposition to resistance. A horse in drawing a cart or 
a plough does a certain amount of work, which depends on the 
resistance and the distance traversed. The force which the 
horse exerts, and the distance through which he moves, may be 
regarded as the two elements of the work done. If r lbs. is the 
constant resistance or force, and x feet the effective distance 



INTEGRATION AS A SUMMATION OF ELEMENTS, 241 



through which the body moves, rx units of work will be done. 
By effective distance is meant the distance measured in the 
same direction as that in which the force is acting. Thus, when 
the resistance is constant, the amount of work may be represented 
by the area of a rectangle whose base is the distance (x) and 
whose altitude is the resistance (r). 

If the resistance or force is a variable dependent on the dis- 
tance x, it may be represented hjf'(x), in which case the amount 
of work may be found by taking the sum of its elements, thus: 
In Fig. 43, if the force fix) (= BP) act through the small 
effective distance h (= BC), the work done will be in excess of 
f'(x)h (= BCDP) only by the acceleration of work {PDP f ) 
during that interval. Hence, /' (x)dx is the differential value of 
the zth element of work. Therefore the quantity of work 
between the. limits x = x x and x = # 2 is, viz. : 

2lWov l ™l^y\x)l l -\=f x y'{x)dx, . (N) 

in which the effective distance is z 2 — x x . 

Cor. I. Effective distance, resistance, and work, and effective 
distance, force, and energy, bear the same relation 
to one another as the abscissa, ordinate, and area of 
a plane curve referred to rectangular co-ordinates, 
respectively. 

Example. Let it be required to compute the 
quantity of work necessary to compress the spiral 
spring of the common spring- balance to any given 
degree, say from AB to DB.* 

Let the resistance ( = /'(#)) vary directly as 
the degree of conrpression, and denote the distance 
AD' by x\ then will 

f'{x) == mx, Fig. 54. 

where m is the resistance of the spring when the balance 




Bartlett's Analytical Mechanics, page 39. 



242 DIFFERENTIAL AND INTEGRAL CALCULUS. 

is compressed through the distance unity. Substituting in (N), 
making x l = and x a — AD, we have 



the work = / mx dx = — 1- G 



= -kmx n 

o 



Cor. I. If m = 10 pounds and x 2 — 3 ft., then will 

the work = 45 units of work; 

that is, the quantity of work will be equal to that required to 
raise 45 pounds through a vertical height of one foot. 

CENTRE OF GRAVITY. 

239. The bodies here considered are supposed to be of uni- 
form density; that is, equal quantities of a body have equal 
weights. 

The Centre of Gravity of a body is a point so situated that 
the force of gravity produces no tendency in the body to rotate 
about any axis passing through this point. 

The Moment of any element or particle of a body with 
reference to any horizontal axis is the product of the magnitude 
or weight of the element by the horizontal distance of its centre 
of gravity from the axis, and measures the tendency of the 
element, under the influence of gravity, to produce rotation 
about the axis. The moment of the body itself is the sum of 
the moments of its elements. If the axis of reference passes 
through the centre of gravity of the body, the moment of the 
body must be zero, otherwise the moments of the elements would 
not neutralize one another, and the body would rotate. 

240. To find the centre of gravity of a plane area. 

In Fig. 43, suppose the plane curve placed in a horizontal 
position, and let A = the area of EGQF, % — OB, y = BP, 
x i — OE, x 2 — OG. Also let {x',y') be the centre of gravity 
of A, and (x -\- a, y -j- /3) be the centre of gravity of the rect- 
angle BCDP (= yh). Evidently the limit of a, as BC or h 
approaches 0, is 0. 



INTEGRATION AS A SUMMATION OF ELEMENTS. 243 

The moment of BCDP with respect to an axis passing 
through (x' ', y') and parallel to the axis of y is (x -j- a — x')y]i, 
which is the measure of the tendency of this rectangle [BCDP) 
to produce rotation about the given axis, and therefore the 
tendency of all the similar rectangles to produce rotation is 

^ \x -\- a — x']yJi. The smaller the rectangles the nearer 

their sum comes to the whole area of the curve, and therefore 
the tendency of A to rotate about the given axis is 

, . ^ X \x + <* — x')yli —I i x — x')ydx: 

h or a _ x l v ' . }J «4, 

but as the axis of reference passes through the centre of gravity 
of A, this must equal zero. 

/ {x — x')ydx — I xydx — x' I ydx = 0. 

xydx -^- / ydx = / xydx ~ A. (P) 
In like manner we find 

y' 



f X 'fdx + A (Q) 



241. To find the centre of gravity of a plane curve- 

In Fig. 44, suppose the plane curve PQ (= s) placed in a 
horizontal position; let x = OB, y = BP, x 2 — OG, and 
h = BC; also let (#', y') be the centre of gravity of s, and 
(x -f- a, y -f- fi) the centre of gravity of the tangent 



The limit of a, as h approaches 0, is evidently 0. 



244 DIFFERENTIAL AND INTEGRAL CALCULUS. 

The moment of Pt with respect to an axis passing through 
(x' t y') and parallel to the axis of y is (%-\-<x—z')a/ l _l (-£) h 
and the sum of the moments of all the tangents like Pt is 



2 



[x + a — x']fl -j- -y^j h. The limitof this sum, as h and a 
approach 0, is equal to / (x — x')ds, which is the moment of 

V X 

limit a ' 2 
s with respect to the given axis, since s = 7 . ^ , [Pt\ 

I *{x — x')ds = I *xds — x' I ds = 0. 

Whence x f — / xds -f- s (R) 

In like manner we find 

y' = pyds + s (S) 

242. Let the student prove in a similar manner that the 
formula for finding the centre of gravity of a solid of revolution 
whose axis is the axis of x is 



x' = n I *xy*dx -+- the volume; 
or x' = I ™xy*dx -i- I *y*dx. 



. . . (T) 



Note. — As the centre of gravity must evidently be on the 
axis of revolution, the formula given above entirely determines 
it. The same is true of the following formula. 

Prove that the formula for finding the centre of gravity of 
any surface of revolution whose axis is the axis of x is 

px<x px^ 

x' — I ocyds 4- / yds (U) 

t/X\ ' VXi 



INTEGRATION AS A SUMMATION OF ELEMENTS. 245 
EXAMPLES. 
1. Determine the centre of gravity (G) of an isosceles 



triangle. 



Let OD, the altitude, = a, DC, half of the base, 

l)X 

= b, OB = x, BP = y; then y = — , and, by for- 




mula (P), 

na j bx 2 

I xv ax J ~dx 

t r\n t/0 * J CI o 

X' — OG — -rj^n ~ FT ~ # a * 

area ADC \ab 

that is, the distance of the centre of gravity from the vertex 
of the triangle is equal to two thirds of the altitude of the 
triangle, 

2. Determine the centre of gravity of the area bounded by 
the parabola y 2 = 4ax and the double ordinate (2y) perpendicular 
to the axis of x. x' = \x, y' — 0. 

3. Find the centre of gravity of the area of the curve 
xy 2 = ¥(a — x). x' = \a, y' = 0. 

4. Find the centre of gravity of the area of the cycloid. 

x' = nr, y' = fr. 

5. Find the centre of gravity of the area of x^ -{- y* = a* 
lying in the first quadrant. , _ , _ 256 a 

X ~ y ~ 315 * 

6. Find the centre of gravity of the area of a 2 y 2 -f b 2 x 2 = a 2 ¥, 
lying in the first quadrant. , _ 4a , _ 4& 

X ~ 3? y ~ 3^* 

7. Find the centre of gravity of the arc of the circle x 2 -j- y 2 
— r 2 lying in the first quadrant. (See formulas (E) and (S).) 

2r 

x' = y' = —. 

* 71 

8. Find the centre of gravity of the arc of the curve x*-\- y* 
= a* lying in the first quadrant. x' = y' — \a. 



246 DIFFERENTIAL AND INTEGRAL CALCULUS. 

9. Find the centre of gravity of the arc of the cycloid 

x = r vers -1 - — V2ry — y' 2 , 

lying between (0, 0) and {nr, 2r). x' = fr, y f = -f- i^*> 

10. Find the centre of gravity of the paraboloid generated 
by revolving y 2 = kmx about the axis of x. (See formula (T).) 

x' — \x. 

11. Find the centre of gravity of the segment of a sphere- 
generated by revolving y 2 = 2rx — x 2 about the axis of x. 

, _ x(8r — 3x) 
4(3r — x) ' 
When x = r, %' = §r. 

12. A semi-ellipsoid is formed by the revolution of a semi- 
ellipse about its major axis; find the distance of the centre of 
gravity of the solid from the centre of the ellipse. x' = § a. 

13. Find the centre of gravity of the convex surface of the 
cone generated by revolving the line y = mx about the axis of x. 
(See formula (U).) x f = §&. 

14. Find the centre of gravity of the surface of a spherical 
segment whose altitude is x. x f = \x. 

15. Find the centre of gravity of the surface of the parabo- 
loid generated by revolving y 2 = Amx about the axis of x. 

, 1 (3z - 2m) (x + mf + 2w* . 
5 (x + ra) 1 — wi 1 



APPENDIX. 



A,. Differentiable Functions. A function, y = f(x), is 

Ay 
said to be differentiable when -~ approaches a definite limit as 

Ax approaches zero. Thus, y = Vx is differentiable, since 
(Art. 10, ex. 5) 

4y _ 1 

Ax |V + A + Vx' . 

approaches the definite limit, — — , as h approaches zero, x f 

4irX 

being any particular or definite value of x from which li is 
estimated. 

All ordinary continuous functions are differentiable, but this 
does not follow from the mere fact that the functions are con- 
tinuous, for there are functions which are continuous and yet 
have no differential coefficients.* Functions of this kind, how- 
ever, are of such rare occurrence that the distinction between 
continuity and differentiability is seldom made in works on the 
Differential Calculus. That is, every function is regarded as 
continuous and differentiable between certain limits. 

Ati 
The limit (m x ) of -p in any particular case can often be 

conveniently determined by assuming that Ay = mji + mji 2 , 

* See Harkness and Morley's Theory of Functions. 

247 



248 APPENDIX, 

which is true of all differentiate functions of a single variable, 
and then finding the value of m l9 as in A 4 . The general 
values of m\ and m 2 , and the exact conditions under which 
Ay = mji -j- mji' 2 holds, are given in A 7 . 

A 2 . Another Illustration of the Formula Ay = mji + mji 2 . 
Suppose that a moving body has traversed a distance (s) in the 
time t, and that the value of s in terms of t is 

*=/(0 (i) 

Suppose we wish to find the actual velocity (v x ) at the end of 
the time /,. Let At, an increment of t estimated from t l9 be 
any arbitrary period immediately succeeding the end of the time 
t l9 then the distance traversed by the body in that period Avill 
be the corresponding increment of s, viz., 

As = f(t + At) - f{t) = m x At + m,(At)\ . . (2) 

The mean velocity (v) of a moving body, during any period 
of time, is the quotient obtained by dividing the distance 
traversed by the body by the length of the period. Therefore 
the mean velocity during the period At is 

^ = m t + m 2 At =.v (3) 

Now this mean velocity evidently approaches the actual 
velocity v x as At approaches 0, indefinitely. Hence taking the 
limit of (3), we have 

limit of —r- = > m 1 = v x . 

As 
That is, the limit of —r, or m 1 , is the actual velocity of the 

body at the end of the time t x , and hence m x At is what As would 
have been had it varied as At or had the actual velocity i\ re- 
mained constant, and m 2 /i 2 is the acceleration of s during the 
period At, 



APPENDIX. 249 

A 3 . The Differentials of Independent Variables are, in 
general, Variables. In differentiating y = f(x) successively 
dx is usually regarded as a constant; that is, as having the same 
value for all values of x. " This hypothesis/' say Eice and John- 
son,* "greatly simplifies the expressions for the second and 
higher differentials of functions of x, inasmuch as it is evidently 
equivalent to making all differentials of x higher than the first 
vanish." Again, "A differential of the second order or of 
a higher order," says Byerly,f " has been defined by the aid of a 
derivative, which always implies the distinction between func- 
tion and variable, and on the hypothesis of an important dif- 
ference in the natures of the increments of function and vari- 
able; namely, that the increment of the independent variable is 
a constant magnitude, and that, consequently, its derivative and 
differential are zero." 

The impressions which these and similar statements in other 
excellent works are likely to make on the mind of the student 
are (a) that all independent variables vary uniformly, and (b) 
that they must vary in this manner in order that the differen- 
tials of their differentials shall be zero. 

That an independent variable may vary uniformly, as in Eate 
of Change, is granted; but differentials in general are variables 
whose limits are zero. Indeed one of the most important and 
essential properties of the differential of an independent vari- 
able is its independent variability. The imposition of any con- 
dition on a group of variables by w T hich they may be expressed in 
terms of one another at once destroys the independence of the 
variables, and this is the case of variables under the hypothesis 
of uniform variation, or rate of change. 

Thus, let u =f(x, y), and let us suppose x and y to vary 
simultaneously and uniformly; then dx = mdt and dy = m'dt; 
whence x = mt -f C and y — m't + C. Eliminating t and 
solving for y, we have y = <f>(x). Hence the supposition of 
uniform* change renders the hypothesis of more than one in- 

* Dif . Calculus, Art. 79. f Dif . Calculus, Art. 204. 



250 APPENDIX. 

dependent variable impossible. Therefore, if independent 
variables (which are supposed to vary simultaneously) must vary 
uniformly in order that their higher differentials shall be zero, 
the successive differentials of u = f(y, x, z) can be obtained 
only by destroying the independence of all the variables x 9 y, z 
except one. 

Hence, in general, if the differential of dx, with respect to 
x, is zero, it is due to the' fact that dx is a variable which is 
independent of x. 

A 4 . To differentiate -a v and log a v independently of Art. 
78. 

Let y = a v , where v is a function of x. 

Increasing x by h, etc., and assuming that a v is differentiable 
(AJ, we have 

Ay , (a^ — 1\ , a 

Av \ Av J * a 

Taking the limits, remembering that m^Av vanishes with h 
and that a v is constant with respect to li, we have 

where m 1 and a v are definite quantities. Therefore the limit of 
- must be a definite quantity (m f say), not zero, and de- 

ZjV 

pendent solely on the base a, since it is evidently independent 
of x and h. 

~j — a v m r , or dy = tfm'dv (1) 

Cor. I. Since m' depends on the base and the base is arbi- 
trary, we may suppose the base to have such a value (say e) that 



APPENDIX. 251 

Coe. II. To find the value of m' in (1), let 

a v = e u (2) 

Differentiating (2), a v m'dv — e u du (3) 

rjii. 
From (2) and (3), m' = ~: .;..... (4) 

Taking log e of (2), v log e a = u (5) 

Differentiating (5), dv log e a = du (6) 

Whence log e a = — (7) 

Equating (4) and (7), m* — log e a. (8) 

d{a v ) = a v log, a dv (9) 

Cok. III. To differentiate y — log a v, we write it under the 
form of 

a y = v (10) 

Differentiating (10), a v log c ady — dv; 

whence dv = , .or log a e — (11) 

J log e a v & v ' 

A 5 . A Rigorous Proof of Taylor's Formula. In what fol- 
lows, the function f(y) and its n successive derivatives are sup- 
posed to be differentiable, and finite and continuous between 
the limits y and y -f x. 

Lemma. If F(z) is continuous bet wen z = a and z = y, and 
if F(a) = F{y) = 0, then F\z), if continuous, must equal zero 
for some value of z between a and y. 

For, as z changes from a to y, F(z) passes from to 0, that 
is, F(z) increases and then decreases, or vice versa; hence F'{z) 
must change from -J- to — or from — to -f- , and therefore, since 
it is continuous, pass through 0. 



252 APPENDIX. 

In what follows 6 will represent a positive proper fraction; 
that is, < 6 < 1. Hence, < dx < x, and, by the lemma, 
F\y + B(a - y)] = 0. 

Under the given hypotheses, we have (A,) 

f n -\y + x) =-f n -\y) + m x x + m<x\ 

or / w -% + ^)=/ n -%) + ^ (1) 

where s (= in 1 -{- mjc) is continnous between y and ?/ -f- x. 

Multiply (1) by dx, regard y constant, and integrate, and we 
have 

f n ~\y + x) =f n - 2 (y) + xf"~\y) + fxs dx, . . (2) 

the constant C being f n ~ 2 (y), since f n ~ 2 (y -\-x) =f*-*(y) when 
x — 0. 

Multiply (2) by dx, and integrate, denoting / / by / , 

and we have 

P'\y -j- x) =f n - s (y) + xf n ~\y) + ^■/"(y) + fxs dx\ (3) 

tf being /-%).' 

Continuing thus, to n — 1 integrations, we have 

/(y + *) = /ty) + *T(y) + £/"(») 

l£ 

+ • • • T,r=T/""'(y) + /""^ 'fe- 1 . (4) 

/n-1 
xs dx 71 ' 1 , 

which is the remainder after n terms. 

In (4) put a — y for x, and s f for the corresponding value of 
s, transpose, and we have 

/(a) -m - ^ny) - ^r^/"w 

- • • • ( -^~rf n -^) - f"~\»-y)s (-dyy-^o. (5) 



APPENDIX. 253 

Let F(z) be a function such that 

-■■• ( V-i " ^"' (z) - /" c*-*^-*)^ • • • (6) 

Since s', having the same value as in (5), is independent of 
z, the last term of (6) is equal to — -j —s'. 

Evidently F(z) = 0, first when z = y by (5), second when 
z = a; and since f{y), f'(y),f"(y), etc., are all continuous from 
y to y -j- x (= a),f(z),f'(z),f"(z), etc. (and therefore F(z) and 
i^'(^) ), are continuous between the same limits. Therefore, by 
the lemma, F'(y -f 6(a - y) ) = 0. 

Differentiating (6) to obtain F'(z), we have 

F'{z) = -f'(z) + /'(*) - fc-=^/"(«j +_irL£/"(») 

That is, *»(,) = - ( "^f '[/•(*) - *']. ... (7) 



Xow substituting y + #(# — y) for 2, observing that for this 
value of z, F\z) = 0, and dividing by -■ - — , we have 



s'=f n [y + V(a-i/)1 (8) 

In (6) substitute this value of s', and then put x for z 



254 APPENDIX. 

(whence F(z) = 0), y -j- x for a (whence a — y = x), transpose, 
and we have 



f(y + i) =f(y) + xf'{y) + jj"{y) 



+ ■■■ u^T-^'W + Tnf n iy + fe )- (9) 



The last term 



-j^/% + &0 = Rn, say, 



is called the remainder in Taylor's formula. It is obtained by 
differentiating f(y) n times and substituting?/ + ^ x ^ or V i n the 
final or nt\i derivative. 

When the function f(y + #) is such that this remainder ap- 
proaches as n approaches co , the series will be convergent, 
otherwise it will be divergent. Hence the remainder enables us 
to ascertain the conditions under which any given function of 
the sum of two variables is developable by Taylor's formula, 
and also to find the limits of the error we make in stopping at 
any term of the series. 

EXAMPLE. 
1. f(y + x)= log (y + x). See Art. 125. 

|f» - 1 

Since f{y) = log y, f n (y) = 

£.= -(- 1) 

v ' n\y -\- uxl 



Now, since < < 1, a— is finite and a proper fraction 

y -\- Ox r r 

if x — or < y. Therefore R n approaches as n approaches co , 

and log (y + x) is developable if x — or < y. 



y 
i 



APPENDIX. 255 

Again, since < 6 < 1, the true numerical value of R n lies 

between I(-V and L(_ ?— )" Hence if - (- l)"-(-V be 

n \yl n\y + xl ' n \y J 

substituted for R n in (1), Art. 125, the series will be the value 

1 / x 
of log (y -f- x) to within less than — ( — — 

lb \'U I Jj J 

A 6 . A Similar Completion of Maclaurin's Formula may be 

obtained by making y = in (9), A & . Thus, 

/(*)=/(o) + z/'(0) + |r/"(o) 

+ --^l/"- I (0)+f/*(^)- CD 

EXAMPLE. 
f(x) = e x . 

Since Mx) = tf" i?„ = ^f-e 9x . 

For any finite value of x 9 (1) the fraction j- evidently ap- 
proaches as n approaches x , and (2), since < 6 < 1, e 6x is 
finite. Hence the limit of E„ , as n approaches oo , is 0, and e x 
is developable, for all finite values of x. 

x n 
Again, as < 6 < 1, the true value of R n lies between -j— 

I? 

x n 
and -, — e x . Therefore the sum of all the terms after the ^th in 

\n 

x n 
series (N), Art. 127, is less than - — e x . 

A 7 . To determine the values of m x and m 2 in the formula 

Ay = mji + mji 2 (Art. 24). 

Since y = f(x), Ay = f(x + 7i) - f(x). 



256 APPENDIX, 

By Taylor's formula, 

f{x + ») =/(x) + hf(x) + ^f"(x + eh). 

Ay = f'(x)h + if"(x + 6h)W. 

Comparing this with the given formula, we have m 1 — f'{x) 
and m 2 = \f"(x + Oh), where < 6 < 1. 

Therefore the formula Ay = mji -f mji 2 is true in reference 
to y — f(x), (1) if f(x) and f'(x) are diiferentiable, and (2) if 
f(x), f'(x), and f"(x) are finite and continuous between the 
limits x and x -f- h. 



